Problems
Section 27.1
The Principle
of Linear
Superposition,
Section 27.2 Young's Double-Slit Experiment
In a Young's double-slit experiment, the wavelength of the light used is 520 nm (in vacuum), and the separation
1.
between the slits is
. Determine the angle that locates
,
(a)the dark fringe for which
Answer:
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(b)the bright fringe for which
Answer:
,
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(c)the dark fringe for which
Answer:
, and
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
(d)the bright fringe for which
Answer:
.
REASONING The angles that determine the locations of the dark and bright fringes in a Young's
double-slit experiment are related to the integers m that identify the fringes, the wavelength of the light,
and the separation d between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
REASONING The angles that determine the locations of the dark and bright fringes in a Young's double-slit
experiment are related to the integers m that identify the fringes, the wavelength of the light, and the separation d
between the slits. Since values are given for m, , and d, the angles can be calculated.
SOLUTION The expressions that specify in terms of m, , and d are as follows:
(27.1)
(27.2)
Applying these expressions gives the answers that we seek.
(a)
(b)
(c)
(d)
2.In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright
fringe is
. Find the ratio
of the slit separation d to the wavelength of the light.
3.Two in-phase sources of waves are separated by a distance of 4.00 m. These sources produce identical waves that
have a wavelength of 5.00 m. On the line between them, there are two places at which the same type of interference
occurs.
(a)Is it constructive or destructive interference, and
Answer:
Destructive interference occurs.
(b)where are the places located?
Answer:
3.25 m and 0.75 m from one of the sources
in a Young's double-slit experiment is located at an angle of
. What is the angle
4.The dark fringe for
that locates the dark fringe for
?
5.
In a Young's double-slit experiment, the seventh dark fringe is located 0.025 m to the side of the central bright
fringe on a flat screen, which is 1.1 m away from the slits. The separation between the slits is
the wavelength of the light being used?
Answer:
. What is
6.
Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is
. The other wavelength is
and is unknown. On a viewing screen, the light with wavelength
produces its third-order bright fringe at the same place where the light with wavelength
produces its
fourth dark fringe. The fringes are counted relative to the central or zeroth-order bright fringe. What is the unknown
wavelength?
7.
In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The
separation between the slits is
. The total width of the screen is 0.20 m. In one version of the setup,
the separation between the double slit and the screen is
, whereas in another version it is
.
On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do
not include the central bright fringe in your counting.
Answer:
6 (version A), 4 (version B)
*8.At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength
625 nm falls on a double slit whose slit separation is
?
In a Young's double-slit experiment the separation y between the second-order bright fringe and the central
bright fringe on a flat screen is 0.0180 m when the light has a wavelength of 425 nm. Assume that the angles that
locate the fringes on the screen are small enough so that
. Find the separation y when the light has a
wavelength of 585 nm.
Answer:
0.0248 m
**10.In Young's experiment a mixture of orange light (611 nm) and blue light (471 nm) shines on the double slit. The
centers of the first-order bright blue fringes lie at the outer edges of a screen that is located 0.500 m away from the
slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or
away from the slits) should the screen be moved so that the centers of the first-order bright orange fringes will just
appear on the screen? It may be assumed that is small, so that sin
.
**11.
A sheet that is made of plastic
covers one slit of a double slit (see the drawing). When the
*9.
double slit is illuminated by monochromatic light
dark rather than bright. What is the minimum thickness of the plastic?
, the center of the screen appears
Problem 11
Answer:
487 nm
REASONING The light that travels through the plastic has a different path length than the light that
passes through the unobstructed slit. Since the center of the screen now appears dark, rather than bright,
destructive interference, rather than constructive interference occurs there. This means that the difference
between the number of wavelengths in the plastic sheet and that in a comparable thickness of air is .
SOLUTION The wavelength of the light in the plastic sheet is given by Equation 27.3 as
The number of wavelengths contained in a plastic sheet of thickness t is
The number of wavelengths contained in an equal thickness of air is
where we have used the fact that
. Destructive interference occurs when the difference,
, in the number of wavelengths is :
Solving this equation for t yields
.
Section
27.3 ThinFilm
Interference
12.You are standing in air and are looking at a flat piece of glass
on which there is a layer of transparent
plastic
. Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated
glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the
thickness of the layer.
13.
A nonreflective coating of magnesium fluoride
covers the glass
of a camera lens.
Assuming that the coating prevents reflection of yellow-green light
the minimum nonzero thickness that the coating can have.
Answer:
102 nm
, determine
REASONING To solve this problem, we must express the condition for destructive interference in terms of
the film thickness t and the wavelength
of the light as it passes through the magnesium fluoride coating. We must
also take into account any phase changes that occur upon reflection.
SOLUTION Since the coating is intended to be nonreflective, its thickness must be chosen so that destructive
interference occurs between waves 1 and 2 in the drawing. For destructive interference, the combined phase difference
between the two waves must be an odd integer number of half wavelengths. The phase change for wave 1 is
equivalent to one-half of a wavelength, since this light travels from a smaller refractive index
larger refractive index
toward a
.
Similarly, there is a phase change when wave 2 reflects from the right surface of the film, since this light also travels
from a smaller refractive index
toward a larger one
. Therefore, a phase change of
one-half wavelength occurs at both boundaries, so the net phase change between waves 1 and 2 due to reflection is
zero. Since wave 2 travels back and forth through the film and, and since the light is assumed to be at nearly normal
incidence, the extra distance traveled by wave 2 compared to wave 1 is twice the film thickness, or . Thus, in this
case, the minimum condition for destructive interference is
The wavelength of light in the coating is
(27.3)
Solving the above expression for t, we find that the minimum thickness that the coating can have is
14.When monochromatic light shines perpendicularly on a soap film
with air on each side, the second
smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 nm. What is the
vacuum wavelength of the light in nm?
15.
A transparent film
film has a thickness that is
this film has been designed?
Answer:
16.
17.
is deposited on a glass plate
to form a nonreflecting coating. The
. What is the longest possible wavelength (in vacuum) of light for which
A tank of gasoline
is open to the air
. A thin film of liquid floats on the gasoline and has
a refractive index that is between 1.00 and 1.40. Light that has a wavelength of 625 nm (in vacuum) shines
perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive
interference. The thickness of the film is 242 nm and is the minimum nonzero thickness for which constructive
interference can occur. What is the refractive index of the film?
Review Conceptual Example 4 before beginning this problem. A soap film with different thicknesses at
different places has an unknown refractive index n and air on both sides. In reflected light it looks multicolored. One
region looks yellow because destructive interference has removed blue
from the reflected light,
while another looks magenta because destructive interference has removed green
. In these
regions the film has the minimum nonzero thickness t required for the destructive interference to occur. Find the ratio
.
Answer:
1.18
*18.
A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the
minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light
. Assuming that the visible spectrum extends from 380 to 750 nm, for which
visible wavelength(s) in vacuum will the film appear bright due to constructive interference?
*19.
Orange light
shines on a soap film
that has air on either side of it. The light
strikes the film perpendicularly. What is the minimum thickness of the film for which constructive interference causes
it to look bright in reflected light?
Answer:
115 nm
REASONING To solve this problem, we must express the condition for constructive interference in terms of
the film thickness t and the wavelength
of the light in the soap film. We must also take into account any phase
changes that occur upon reflection.
SOLUTION For the reflection at the top film surface, the light travels from air, where the refractive index is smaller
, toward the film, where the refractive index is larger
. Associated with this reflection there is
a phase change that is equivalent to one-half of a wavelength. For the reflection at the bottom film surface, the light
travels from the film, where the refractive index is larger
, toward air, where the refractive index is
smaller
. Associated with this reflection, there is no phase change. As a result of these two reflections,
there is a net phase change that is equivalent to one-half of a wavelength. To obtain the condition for constructive
interference, this net phase change must be added to the phase change that arises because of the film thickness t,
which is traversed twice by the light that penetrates it. For constructive interference we find that
or
Equation 27.3 indicates that
. Using this expression and the fact that
thickness t, we find that the condition for constructive interference becomes
for the minimum
or
*20.
The drawing shows a cross section of a planoconcave lens resting on a flat glass plate. (A planoconcave lens
has one surface that is a plane and the other that is concave spherical.) The thickness t is
. The lens is
illuminated with monochromatic light
, and a series of concentric bright and dark rings is
formed, much like Newton's rings. How many bright rings are there?
(Hint: The cross section shown in the drawing reveals that a kind of air wedge exists between the place where the two
pieces of glass touch and the top of the curved surface where the distance t is marked.)
**21.A piece of curved glass has a radius of curvature of
and is used to form Newton's rings, as in Figure
27.13. Not counting the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at
the outer edge of the curved piece of glass. The light being used has a wavelength of 654 nm in vacuum. What is the
radius R of the outermost dark ring in the pattern?
(Hint: Note that r is much greater than R, and you may assume that tan
for small angles, where must be
expressed in radians.)
Answer:
0.0256 m
**22.
A uniform layer of water
lies on a glass plate
. Light shines perpendicularly on
the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the
light is 432 nm in vacuum and also when it is 648 nm in vacuum.
(a)Obtain the minimum thickness of the film.
(b)Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to
750 nm, determine the visible wavelength(s) in vacuum for which the film appears completely dark.
Section
27.5
Diffraction
23.
(a)As Section 17.3 discusses, high-frequency sound waves exhibit less diffraction than low-frequency sound
waves do. However, even high-frequency sound waves exhibit much more diffraction under normal
circumstances than do light waves that pass through the same opening. The highest frequency that a healthy
ear can typically hear is
. Assume that a sound wave with this frequency travels at 343 m/s
and passes through a doorway that has a width of 0.91 m. Determine the angle that locates the first
minimum to either side of the central maximum in the diffraction pattern for the sound. This minimum is
equivalent to the first dark fringe in a single-slit diffraction pattern for light.
Answer:
(b)Suppose that yellow light
passes through a doorway and that the first
dark fringe in its diffraction pattern is located at the angle determined in part (a). How wide would this
hypothetical doorway have to be?
Answer:
24.A dark fringe in the diffraction pattern of a single slit is located at an angle of
. With the same light, the
same dark fringe formed with another single slit is at an angle of
. Find the ratio
of the widths of
the two slits.
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 675 nm.
25.
Determine the angle that locates the first dark fringe when the width of the slit is
(a)
and
Answer:
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a)
When the slit width is
and
, we find, according to
Equation 27.4,
(b)
Similarly, when the slit width is
(b)
and
, we find
.
Answer:
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a)
When the slit width is
and
, we find, according to
Equation 27.4,
(b)
Similarly, when the slit width is
and
, we find
REASONING This problem can be solved by using Equation 27.4 for the value of the angle when
(first dark fringe).
SOLUTION
(a)
When the slit width is
and
, we find, according to Equation
27.4,
(b)
Similarly, when the slit width is
and
, we find
26.
27.
A slit has a width of
. When light with a wavelength of
passes through this
slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the
same place, this slit is replaced with a second slit (width
), and a wavelength of
is used. The width of
the central bright fringe on the screen is observed to be unchanged. Find
.
Light that has a wavelength of 668 nm passes through a slit
wide and falls on a screen that is 1.85
m away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either
side?
Answer:
0.576 m
REASONING The drawing shows a top view of the slit and screen, as well as the position of the central
bright fringe and the third dark fringe. The distance y can be obtained from the tangent function as
tan . Since
L is given, we need to find the angle before y can be determined. According to Equation 27.4, the angle is related to
the wavelength of the light and the width W of the slit by
, where
since we are interested in
the angle for the third dark fringe.
SOLUTION We will first compute the angle between the central bright fringe and the third dark fringe using Equation
27.4 (with
):
The vertical distance is
28.
*29.
Light shines through a single slit whose width is
. A diffraction pattern is formed on a flat screen
located 4.0 m away. The distance between the middle of the central bright fringe and the first dark fringe is 3.5 mm.
What is the wavelength of the light?
Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit
whose width is
and strike a screen 1.20 m from the slit. Two diffraction patterns are formed on the
screen. What is the distance (in cm) between the common center of the diffraction patterns and the first occurrence of
a dark fringe from one pattern falling on top of a dark fringe from the other pattern?
Answer:
3.18 cm
*30.
*31.
The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen
and the slit. Find the ratio
of the wavelength of the light to the width W of the slit.
How many dark fringes will be produced on either side of the central maximum if light
incident on a single slit that is
Answer:
8
wide?
is
REASONING The angle that specifies the location of the
dark fringe is given by
(Equation 27.4), where is the wavelength of the light and W is the width of the slit. When has its maximum value of
, the number of dark fringes that can be produced is a maximum. We will use this information to obtain a value
for this number.
SOLUTION Solving Equation 27.4 for m, and setting
, we have
Therefore, the number of dark fringes is
.
**32.In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18 000 times farther
from the slit than the slit is wide. What is the ratio
, where is the wavelength of the light shining through the
slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that
.
Section 27.6 Resolving Power
33.
Two stars are
apart and are equally distant from the earth. A telescope has an objective lens with a
diameter of 1.02 m and just detects these stars as separate objects. Assume that light of wavelength 550 nm is being
observed. Also assume that diffraction effects, rather than atmospheric turbulence, limit the resolving power of the
telescope. Find the maximum distance that these stars could be from the earth.
Answer:
34.
It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward
home plate. One way to judge this claim is to estimate the distance at which a batter can first hope to resolve two
points on opposite sides of a baseball, which has a diameter of 0.0738 m.
(a)Estimate this distance, assuming that the pupil of the eye has a diameter of 2.0 mm and the wavelength of
the light is 550 nm in vacuum.
(b)Considering that the distance between the pitcher's mound and home plate is 18.4 m, can you rule out the
claim based on your answer to part (a)?
35.
Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the
pupils of your eyes have diameters of about 7.0 mm. The taillightsof this car are separated by a distance of 1.2 m and
emit red light
. How far away from you is this car when its taillights appear to
merge into a single spot of light because of the effects of diffraction?
Answer:
REASONING According to Rayleigh's criterion, the two taillights must be separated by a distance s sufficient
to subtend an angle
at the pupil of the observer's eye. Recalling that this angle must be expressed in
radians, we relate
to the distances s and L.
SOLUTION The wavelength is 660 nm. Therefore, we have from Equation 27.6
According to Equation 8.1, the distance L between the observer and the taillights is
36.
An inkjet printer uses tiny dots of red, green, and blue ink to produce an image. Assume that the dot separation on
the printed page is the same for all colors. At normal viewing distances, the eye does not resolve the individual dots,
regardless of color, so that the image has a normal look. The wavelengths for red, green, and blue are
,
, and
. The diameter of the pupil through which light enters the eye is 2.0 mm.
For a viewing distance of 0.40 m, what is the maximum allowable dot separation?
37.
A hunter who is a bit of a braggart claims that from a distance of 1.6 km he can selectively shoot either of
two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do
this without the aid of a telescopic sight on his rifle.
(a)Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the
squirrels as separate objects. In this calculation use a wavelength of 498 nm (in vacuum) for the light.
Answer:
9.7 mm
(b)State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim,
consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to 8
mm, the larger values coming into play as the lighting becomes darker. Note also that under dark conditions,
the eye is most sensitive to a wavelength of 498 nm.
Answer:
The hunter's claim is not reasonable.
38.
Review Conceptual Example 8 as background for this problem. In addition to the data given there, assume that
the dots in the painting are separated by 1.5 mm and that the wavelength of the light is
. Find the
distance at which the dots can just be resolved by
(a)the eye and
(b)the camera.
39.
Astronomers have discovered a planetary system orbiting the star Upsilon Andromedae, which is at a distance
of
from the earth. One planet is believed to be located at a distance of
from the star. Using
visible light with a vacuum wavelength of 550 nm, what is the minimum necessary aperture diameter that a telescope
must have so that it can resolve the planet and the star?
Answer:
2.3 m
*40.
The pupil of an eagle's eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a
distance of 176 m, the eagle sees them as one unresolved object and dives toward them at a speed of 17 m/s. Assume
that the eagle's eye detects light that has a wavelength of 550 nm in vacuum. How much time passes until the eagle
sees the mice as separate objects?
*41.
Consult Multiple-Concept Example 7 to see a model for solving this kind of problem. You are
using a microscope to examine a blood sample. Recall from Section 26.12 that the sample should be placed just
outside the focal point of the objective lens of the microscope.
(a)If the specimen is being illuminated with light of wavelength and the diameter of the objective
equals its focal length, determine the closest distance between two blood cells that can just be
resolved. Express your answer in terms of .
Answer:
REASONING Assuming that the angle
is given by
is small, the distance y between the blood cells
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal
length of the objective). However, the minimum angular separation
of the cells is given by the
Rayleigh criterion as
(Equation 27.6), where is the wavelength of the light and D
is the diameter of the objective. These two relations can be used to find an expression for y in terms
of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that
, we see that
.
(b)Because y is proportional to , the wavelength must be
to resolve cells that are closer
together.
(b)Based on your answer to (a), should you use light with a longer wavelength or a shorter wavelength if
you wish to resolve two blood cells that are even closer together?
Answer:
shorter wavelength
REASONING Assuming that the angle
is given by
is small, the distance y between the blood cells
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal
length of the objective). However, the minimum angular separation
of the cells is given by the
Rayleigh criterion as
(Equation 27.6), where is the wavelength of the light and D
is the diameter of the objective. These two relations can be used to find an expression for y in terms
of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that
, we see that
.
(b)Because y is proportional to , the wavelength must be
together.
REASONING Assuming that the angle
to resolve cells that are closer
is small, the distance y between the blood cells is given by
(8.1)
where f is the distance between the microscope objective and the cells (which is given as the focal length of the
objective). However, the minimum angular separation
of the cells is given by the Rayleigh criterion as
(Equation 27.6), where is the wavelength of the light and D is the diameter of the
objective. These two relations can be used to find an expression for y in terms of .
SOLUTION
(a) Substituting Equation 27.6 into Equation 8.1 yields
Since it is given that
, we see that
.
(b)Because y is proportional to , the wavelength must be
to resolve cells that are closer together.
**42.Two concentric circles of light emit light whose wavelength is 555 nm. The larger circle has a radius of 4.0 cm,
and the smaller circle has a radius of 1.0 cm. When taking a picture of these lighted circles, a camera admits
light through an aperture whose diameter is 12.5 mm. What is the maximum distance at which the camera can
(a)distinguish one circle from the other and
(b)reveal that the inner circle is a circle of light rather than a solid disk of light?
Section
27.7 The
Diffraction
Grating,
Section 27.8 Compact Discs, Digital Video Discs, and the Use of Interference
43.
A diffraction grating is 1.50 cm wide and contains 2400 lines. When used with light of a certain wavelength, a
third-order maximum is formed at an angle of
. What is the wavelength (in nm)?
Answer:
644 nm
REASONING The angle that specifies the third-order maximum of a diffraction grating is
(Equation 27.7), where
is the wavelength of the light, and d is the separation between the slits of the
grating. The separation is equal to the width of the grating (1.50 cm) divided by the number of lines (2400).
SOLUTION Solving Equation 27.7 for the wavelength, we obtain
44.The light shining on a diffraction grating has a wavelength of 495 nm (in vacuum). The grating produces a secondorder bright fringe whose position is defined by an angle of
. How many lines per centimeter does the grating
have?
. For an unknown
45.For a wavelength of 420 nm, a diffraction grating produces a bright fringe at an angle of
wavelength, the same grating produces a bright fringe at an angle of
. In both cases the bright fringes are of the
same order m. What is the unknown wavelength?
Answer:
630 nm
46.
Two diffraction gratings, A and B, are located at the same distance from the observation screens. Light with the
same wavelength is used for each. The separation between adjacent principal maxima for grating A is 2.7 cm, and
for grating B it is 3.2 cm. Grating A has 2000 lines per meter. How many lines per meter does grating B have?
(Hint: The diffraction angles are small enough that the approximation
can be used.)
The wavelength of the laser beam used in a compact disc player is 780 nm. Suppose that a diffraction grating
47.
produces first-order tracking beams that are 1.2 mm apart at a distance of 3.0 mm from the grating. Estimate the
spacing between the slits of the grating.
Answer:
REASONING AND SOLUTION The geometry of the situation is shown below.
From the geometry, we have
Then, solving Equation 27.7 with
for the separation d between the slits, we have
. What is the angle for
48.The first-order principle maximum produced by a grating is located at an angle of
the third-order maximum with the same light?
*49.
A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at
. The
grating is used with light that contains all wavelengths between 410 and 660 nm. What is (are) the wavelength(s) of
the incident light that could have produced this maximum?
Answer:
640 nm and 480 nm
*50.
Light of wavelength 410 nm (in vacuum) is incident on a diffraction grating that has a slit separation of
. The distance between the grating and the viewing screen is 0.15 m. A diffraction pattern is produced
on the screen that consists of a central bright fringe and higher-order bright fringes (see the drawing).
Problem 50
(a)Determine the distance y from the
central bright fringe to the secondorder bright fringe.
(Hint: The diffraction angles are
small enough that the approximation
can be used.)
(b)If the entire apparatus is submerged in water
*51.
, what is the distance y?
Violet light
and red light
lie at opposite ends of the visible
spectrum.
(a)For each wavelength, find the angle that locates the first-order maximum produced by a grating with 3300
lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion
between the two angles. Repeat the calculation above for
Answer:
violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7,
, with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order,
therefore, for violet light,
Similarly for red light,
(b)Repeating the calculation for the second order maximum
(c) Repeating the calculation for the third order maximum
, we find that
, we find that
(d)Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(b)the second-order maximum and
Answer:
violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7,
, with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order,
therefore, for violet light,
Similarly for red light,
(b)Repeating the calculation for the second order maximum
, we find that
(c) Repeating the calculation for the third order maximum
, we find that
(d)Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(c)the third-order maximum.
Answer:
violet light:
red light:
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7,
, with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order,
therefore, for violet light,
Similarly for red light,
(b)Repeating the calculation for the second order maximum
(c) Repeating the calculation for the third order maximum
, we find that
, we find that
(d)Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
(d)From your results, decide whether there is an overlap between any of the “rainbows” and, if so, specify
which orders overlap.
Answer:
The second and third orders overlap.
REASONING The angle that locates the first-order maximum produced by a grating with 3300
lines/cm is given by Equation 27.7,
, with the order of the fringes given by
Any two of the diffraction patterns will overlap when their angular positions are the
same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order,
therefore, for violet light,
Similarly for red light,
(b)Repeating the calculation for the second order maximum
(c) Repeating the calculation for the third order maximum
, we find that
, we find that
(d)Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
REASONING The angle that locates the first-order maximum produced by a grating with 3300 lines/cm is
given by Equation 27.7,
, with the order of the fringes given by
Any two of the
diffraction patterns will overlap when their angular positions are the same.
SOLUTION Since the grating has 3300 lines/cm, we have
(a) In first order,
therefore, for violet light,
Similarly for red light,
(b)Repeating the calculation for the second order maximum
, we find that
(c) Repeating the calculation for the third order maximum
, we find that
(d)Comparisons of the values for calculated in parts (a), (b) and (c) show that the
.
**52.
The distance between adjacent slits of a certain diffraction grating is
. The grating is illuminated by
monochromatic light with a wavelength of 656.0 nm, and is then heated so that its temperature increases by
. Determine the change in the angle of the seventh-order principal maximum that occurs as a result of the
thermal expansion of the grating. The coefficient of linear expansion for the diffraction grating is
. Be sure to include the proper algebraic sign with your answer:
if
the angle decreases.
**53.Two gratings A and B have slit separations
and
, respectively. They are used with the same light and the same
observation screen. When grating A is replaced with grating B, it is observed that the first-order maximum of A is
exactly replaced by the second-order maximum of B.
(a)Determine the ratio
of the spacings between the slits of the gratings.
Answer:
2
(b)Find the next two principal maxima of grating A and the principal maxima of B that exactly replace them
when the gratings are switched. Identify these maxima by their order numbers.
Answer:
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Problems
Section 28.3 The Relativity of Time: Time Dilation
A particle known as a pion lives for a short time before breaking apart into other particles. Suppose that a pion is
moving at a speed of 0.990c, and an observer who is stationary in a laboratory measures the pion's lifetime to be
1.
.
(a)What is the lifetime according to a hypothetical person who is riding along with the pion?
Answer:
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking
apart). Imagine a reference frame attached to the pion, so the pion is stationary relative to this
reference frame. To a hypothetical person who is at rest with respect to this reference frame, these
two events occur at the same place, namely, at the place where the pion is located. Thus, this
hypothetical person measures the proper time interval
for the decay of the pion. On the other
hand, the person standing in the laboratory sees the two events occurring at different locations, since
the pion is moving relative to that person. The laboratory person, therefore, measures a dilated time
interval
. The relation between these two time intervals is given by
(Equation 28.1).
(b)According to the hypothetical person who is at rest in the reference frame attached to the moving
pion, the distance x that the laboratory travels before the pion breaks apart is equal to the speed v of
the laboratory relative to the pion times the proper time interval
, or
. The speed of the
laboratory relative to the pion is the same as the speed of the pion relative to the laboratory, namely,
.
SOLUTION
(a) The proper time interval is
(28.1)
(b)The distance x that the laboratory travels before the pion breaks apart, as measured by the
hypothetical person, is
(b)According to this hypothetical person, how far does the laboratory move before the pion breaks apart?
Answer:
1.5 m
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking
apart). Imagine a reference frame attached to the pion, so the pion is stationary relative to this
reference frame. To a hypothetical person who is at rest with respect to this reference frame, these
two events occur at the same place, namely, at the place where the pion is located. Thus, this
hypothetical person measures the proper time interval
for the decay of the pion. On the other
hand, the person standing in the laboratory sees the two events occurring at different locations, since
the pion is moving relative to that person. The laboratory person, therefore, measures a dilated time
interval
. The relation between these two time intervals is given by
(Equation 28.1).
(b)According to the hypothetical person who is at rest in the reference frame attached to the moving
pion, the distance x that the laboratory travels before the pion breaks apart is equal to the speed v of
the laboratory relative to the pion times the proper time interval
, or
. The speed of the
laboratory relative to the pion is the same as the speed of the pion relative to the laboratory, namely,
.
SOLUTION
(a) The proper time interval is
(28.1)
(b)The distance x that the laboratory travels before the pion breaks apart, as measured by the
hypothetical person, is
REASONING
(a) The two events in this problem are the creation of the pion and its subsequent decay (or breaking apart).
Imagine a reference frame attached to the pion, so the pion is stationary relative to this reference frame. To a
hypothetical person who is at rest with respect to this reference frame, these two events occur at the same place,
namely, at the place where the pion is located. Thus, this hypothetical person measures the proper time interval
for the decay of the pion. On the other hand, the person standing in the laboratory sees the two events
occurring at different locations, since the pion is moving relative to that person. The laboratory person,
therefore, measures a dilated time interval
. The relation between these two time intervals is given by
(Equation 28.1).
(b)According to the hypothetical person who is at rest in the reference frame attached to the moving pion, the
distance x that the laboratory travels before the pion breaks apart is equal to the speed v of the laboratory
relative to the pion times the proper time interval
, or
. The speed of the laboratory relative to
the pion is the same as the speed of the pion relative to the laboratory, namely,
.
SOLUTION
(a) The proper time interval is
(28.1)
(b)The distance x that the laboratory travels before the pion breaks apart, as measured by the hypothetical person,
is
2.A radar antenna is rotating and makes one revolution every 25 s, as measured on earth. However, instruments on a
spaceship moving with respect to the earth at a speed v measure that the antenna makes one revolution every 42 s.
What is the ratio
of the speed v to the speed c of light in a vacuum?
Suppose that you are planning a trip in which a spacecraft is to travel at a constant velocity for exactly six
3.
months, as measured by a clock on board the spacecraft, and then return home at the same speed. Upon your return,
the people on earth will have advanced exactly one hundred years into the future. According to special relativity, how
fast must you travel? Express your answer to five significant figures as a multiple of c—for example, 0.955 85c.
Answer:
REASONING The total time for the trip is one year. This time is the proper time interval
, because it is
measured by an observer (the astronaut) who is at rest relative to the beginning and ending events (the times when the
trip started and ended) and who sees them at the same location in spacecraft. On the other hand, the astronaut
measures the clocks on earth to run at the dilated time interval
, which is the time interval of one hundred years.
The relation between the two time intervals is given by Equation 28.1, which can be used to find the speed of the
spacecraft.
SOLUTION The dilated time interval
is related to the proper time interval
Solving this equation for the speed v of the spacecraft yields
by
.
(28.1)
4.
Suppose that you are traveling on board a spacecraft that is moving with respect to the earth at a speed of 0.975c.
You are breathing at a rate of 8.0 breaths per minute. As monitored on earth, what is your breathing rate?
*5.
A 6.00-kg object oscillates back and forth at the end of a spring whose spring constant is 76.0 N/m. An
observer is traveling at a speed of
measure for the period of oscillation?
Answer:
2.28 s
*6.
relative to the fixed end of the spring. What does this observer
A spaceship travels at a constant speed from earth to a planet orbiting another star. When the spacecraft
arrives, 12 years have elapsed on earth, and 9.2 years have elapsed on board the ship. How far away (in meters) is the
planet, according to observers on earth?
**7.As observed on earth, a certain type of bacterium is known to double in number every 24.0 hours. Two
cultures of these bacteria are prepared, each consisting initially of one bacterium. One culture is left on earth
and the other placed on a rocket that travels at a speed of 0.866c relative to the earth. At a time when the
earthbound culture has grown to 256 bacteria, how many bacteria are in the culture on the rocket, according to
an earth-based observer?
Answer:
16
Section
28.4 The
Relativity
of Length:
Length
Contraction
8.
Suppose the straight-line distance between New York and San Francisco is
(neglecting the curvature of
the earth). A UFO is flying between these two cities at a speed of
relative to the earth. What do the voyagers
aboard the UFO measure for this distance?
How fast must a meter stick be moving if its length is observed to shrink to one-half of a meter?
9.
Answer:
REASONING All standard meter sticks at rest have a length of 1.00 m for observers who are at rest with
respect to them. Thus, 1.00 m is the proper length of the meter stick. When the meter stick moves with speed v
relative to an earth-observer, its length
will be a contracted length. Since both and L are known, v can
be found directly from Equation 28.2,
SOLUTION Solving Equation 28.2 for v, we find that
10.
.
The distance from earth to the center of our galaxy is about 23 000 ly
, as measured by an earth-based observer. A spaceship is to make this
journey at a speed of 0.9990c. According to a clock on board the spaceship, how long will it take to make the trip?
11.
Express your answer in years
.
A tourist is walking at a speed of
along a 9.0-km path that follows an old canal. If the speed of light in
a vacuum were
, how long would the path be, according to the tourist?
Answer:
8.1 km
REASONING The tourist is moving at a speed of
with respect to the path and, therefore,
measures a contracted length L instead of the proper length of
. The contracted length is given by the
length-contraction equation, Equation 28.2.
SOLUTION According to the length-contraction equation, the tourist measures a length that is
12.
A Martian leaves Mars in a spaceship that is heading to Venus. On the way, the spaceship passes earth with a
speed
relative to it. Assume that the three planets do not move relative to each other during the trip. The
distance between Mars and Venus is
, as measured by a person on earth.
(a)What does the Martian measure for the distance between Mars and Venus?
(b)What is the time of the trip (in seconds) as measured by the Martian?
13.Two spaceships A and B are exploring a new planet. Relative to this planet, spaceship A has a speed of 0.60c, and
spaceship B has a speed of 0.80c. What is the ratio
of the values for the planet's diameter that each spaceship
measures in a direction that is parallel to its motion?
Answer:
1.3
14.An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.990c. Relative to a
stationary reference frame fixed to the laboratory, the particle travels a distance of
disintegrating. What are
before
(a)the proper distance and
(b)the distance measured by a hypothetical person traveling with the particle? Determine the particle's
(c)proper lifetime and
(d)its dilated lifetime.
*15.As the drawing shows, a carpenter on a space station has constructed a
ramp. A rocket moves past the space
station with a relative speed of 0.730c in a direction parallel to side . What does a person aboard the rocket measure
for the angle of the ramp?
Answer:
**16.An object is made of glass and has the shape of a cube 0.11 m on a side, according to an observer at rest relative to it.
However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is
3.2 kg determines its density to be
, which is much greater than the density of glass. What is the moving
observer's speed (in units of c) relative to the cube?
**17.A rectangle has the dimensions of
when viewed by someone at rest with respect to it. When you move
past the rectangle along one of its sides, the rectangle looks like a square. What dimensions do you observe when you
move at the same speed along the adjacent side of the rectangle?
Answer:
Section 28.5 Relativistic Momentum
18.At what speed is the magnitude of the relativistic momentum of a particle three times the magnitude of the
nonrelativistic momentum?
19.
What is the magnitude of the relativistic momentum of a proton with a relativistic total energy of
Answer:
?
20.
A spacecraft has a nonrelativistic (or classical) momentum whose magnitude is
. The
spacecraft moves at such a speed that the pilot measures the proper time interval between two events to be one-half
the dilated time interval. Find the relativistic momentum of the spacecraft.
A woman is 1.6 m tall and has a mass of 55 kg. She moves past an observer with the direction of the motion
21.
parallel to her height. The observer measures her relativistic momentum to have a magnitude of
. What does the observer measure for her height?
Answer:
1.0 m
REASONING The height of the woman as measured by the observer is given by Equation 28.2 as
, where is her proper height. In order to use this equation, we must determine the speed v of
the woman relative to the observer. We are given the magnitude of her relativistic momentum, so we can determine v
from p.
SOLUTION According to Equation 28.3
we have
, so
Squaring both sides,
Solving for v and substituting values, we have
Then, the height that the observer measures for the woman is
22.
Three particles are listed in the table. The mass and speed of each particle are given as multiples of the
variables m and v, which have the values
vacuum is
and
. The speed of light in a
. Determine the momentum for each particle according to special relativity.
Particle Mass Speed
a
m
v
b
c
*23.
Starting from rest, two skaters push off against each other on smooth level ice, where friction is
negligible. One is a woman and one is a man. The woman moves away with a velocity of
relative
to the ice. The mass of the woman is 54 kg, and the mass of the man is 88 kg. Assuming that the speed of light
is
, so that the relativistic momentum must be used, find the recoil velocity of the man relative to the
ice.
(Hint: This problem is similar to Example 6 in Chapter 7.)
Answer:
REASONING The magnitude p of the relativistic momentum of an object is given by
(Equation 28.3), where m is the object's mass, v is the object's speed, and c is the speed
of light in a vacuum. The principle of conservation of linear momentum (see Section 7.2) states that the total
momentum of a system is conserved when no net external force acts on the system. This principle applies at
speeds approaching the speed of light in a vacuum, provided that Equation 28.3 is used for the individual
momenta of the objects that comprise the system.
SOLUTION The total momentum of the man/woman system is conserved, since friction is negligible, so that
no net external force acts on the system. Therefore, the final total momentum
must equal the initial
total momentum, which is zero. As a result,
where Equation 28.3 must be used for the momenta
and
. Thus, we find
(1)
We know that
,
, and
. Remember that c has the hypothetical
value of 3.0 m/s. Solving Equation 1 for
reveals that
. We choose the negative value,
since the man and woman recoil from one another and it is stated that the woman moves away in the positive
direction. Therefore, we find that
.
Section 28.6
The
Equivalence
of Mass and
Energy
24.Radium is a radioactive element whose nucleus emits an particle (a helium nucleus) with a kinetic energy of about
25.
(4.9 MeV). To what amount of mass is this energy equivalent?
How much work must be done on an electron to accelerate it from rest to a speed of 0.990c?
Answer:
REASONING According to the work-energy theorem, Equation 6.3, the work that must be done on the
electron to accelerate it from rest to a speed of
is equal to the kinetic energy of the electron when it is moving
at
.
SOLUTION Using Equation 28.6, we find that
26.Review Conceptual Example 9 for background pertinent to this problem. Suppose that the speed of light in a vacuum
were one million times smaller than its actual value, so that
. The spring constant of a spring is
850 N/m. Determine how far you would have to compress the spring from its equilibrium length in order to increase
its mass by 0.010 g.
27.
28.
Suppose that one gallon of gasoline produces
of energy, and this energy is sufficient to operate a car for
twenty miles. An aspirin tablet has a mass of 325 mg. If the aspirin could be converted completely into thermal
energy, how many miles could the car go on a single tablet?
Answer:
Two kilograms of water are changed
into liquid water at
and
(a)from ice at
into steam at
(b)from liquid water at
the water.
. For each situation, determine the change in mass of
29.
Determine the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy
particle has a speed of
(a)
and
Answer:
1.0
when a
REASONING AND SOLUTION
(or less), the relativistic kinetic
(a) In Section 28.6 it is shown that when the speed of a particle is
energy becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here
is
, the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is
.
(b)Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy,
, we find that
(b)0.970c.
Answer:
6.6
REASONING AND SOLUTION
(or less), the relativistic kinetic
(a) In Section 28.6 it is shown that when the speed of a particle is
energy becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here
is
, the ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is
.
(b)Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy,
, we find that
REASONING AND SOLUTION
(or less), the relativistic kinetic energy
(a) In Section 28.6 it is shown that when the speed of a particle is
becomes nearly equal to the nonrelativistic kinetic energy. Since the speed of the particle here is
, the
ratio of the relativistic kinetic energy to the nonrelativistic kinetic energy is
.
(b)
Taking the ratio of the relativistic kinetic energy, Equation 28.6, to the nonrelativistic kinetic energy,
we find that
,
30.Multiple-Concept Example 6 reviews the principles that play a role in this problem. A nuclear power reactor generates
of power. In one year, what is the change in the mass of the nuclear fuel due to the energy being taken
from the reactor?
*31.
Multiple-Concept Example 6 explores the approach taken in problems such as this one. Quasars are believed to
be the nuclei of galaxies in the early stages of their formation. Suppose that a quasar radiates electromagnetic energy
at the rate of
Answer:
*32.
. At what rate (in kg/s) is the quasar losing mass as a result of this radiation?
An electron is accelerated from rest through a potential difference that has a magnitude of
mass of the electron is
. The
, and the negative charge of the electron has a magnitude of
.
(a)What is the relativistic kinetic energy (in joules) of the electron?
(b)What is the speed of the electron? Express your answer as a multiple of c, the speed of light in a vacuum.
*33.
An object has a total energy of
the object's relativistic momentum?
Answer:
and a kinetic energy of
. What is the magnitude of
Section 28.7 The Relativistic Addition of Velocities
34.
You are driving down a two-lane country road, and a truck in the opposite lane is traveling toward you. Suppose
that the speed of light in a vacuum is
. Determine the speed of the truck relative to you when
and the truck's speed is
and
(a)your speed is
and the truck's speed is
. The speeds given in parts (a) and (b) are relative
(b)your speed is
to the ground.
A spacecraft approaching the earth launches an exploration vehicle. After the launch, an observer on earth sees
35.
the spacecraft approaching at a speed of 0.50c and the exploration vehicle approaching at a speed of
. What is
the speed of the exploration vehicle relative to the spaceship?
Answer:
REASONING Let's define the following relative velocities, assuming that the spaceship and exploration
vehicle are moving in the positive direction.
of Exploration vehicle relative to the Spaceship.
of Exploration vehicle relative to an Observer on
of Spaceship relative to an Observer on
The velocity
can be determined from the velocity-addition formula, Equation 28.8:
The velocity
of the observer on earth relative to the spaceship is not given. However, we know that
is the
negative of
, so
.
SOLUTION The velocity of the exploration vehicle relative to the spaceship is
The speed of the exploration vehicle relative to the spaceship is the magnitude of this result or
36.
.
Spaceships of the future may be powered by ion-propulsion engines in which ions are ejected from the back of
the ship to drive it forward. In one such engine the ions are to be ejected with a speed of 0.80c relative to the
spaceship. The spaceship is traveling away from the earth at a speed of
relative to the earth. What is the velocity
of the ions relative to the earth? Assume that the direction in which the spaceship is traveling is the positive direction,
and be sure to assign the correct plus or minus signs to the velocities.
37.The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to
be
. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along
the same line. The velocity of Enterprise 2 relative to Enterprise 1 is
. What is the velocity of Enterprise 2,
as measured by the earth-based observer?
Answer:
*38.
A person on earth notices a rocket approaching from the right at a speed of 0.75c and another rocket
approaching from the left at 0.65c. What is the relative speed between the two rockets, as measured by a passenger on
one of them?
*39.
Refer to Conceptual Example 11 as an aid in solving this problem. An intergalactic cruiser has two types of
guns: a photon cannon that fires a beam of laser light and an ion gun that shoots ions at a velocity of 0.950c relative to
the cruiser. The cruiser closes in on an alien spacecraft at a velocity of 0.800c relative to this spacecraft. The captain
fires both types of guns. At what velocity do the aliens see
(a)the laser light and
Answer:
These answers assume that the direction of the cruiser, the ions, and the laser light is the positive direction.
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach
.
(b)To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d)The aliens see the ions moving away from the cruiser at a velocity
(b)the ions approach them? At what velocity do the aliens see
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach
.
(b)To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d)The aliens see the ions moving away from the cruiser at a velocity
(c)the laser light and
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach
.
(b)To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d)The aliens see the ions moving away from the cruiser at a velocity
(d)the ions move away from the cruiser?
Answer:
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic
cruiser, the ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be
c, regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile
spacecraft see the photons of the laser approach
.
(b)To find the velocity of the ions relative to the aliens, we define the relative velocities as follows:
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions
relative to the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d)The aliens see the ions moving away from the cruiser at a velocity
REASONING AND SOLUTION In all parts of this problem, the direction of the intergalactic cruiser, the
ions, and the laser light is taken to be the positive direction.
(a) According to the second postulate of special relativity, all observers measure the speed of light to be c,
regardless of their velocities relative to each other. Therefore, the aliens aboard the hostile spacecraft see the
photons of the laser approach
.
To
find
the
velocity
of
the
ions
relative
to
the
aliens,
we
define the relative velocities as follows:
(b)
of the Ions relative to the alien Spacecraft
of the Ions relative to the intergalactic
of the intergalactic Cruiser relative to the alien
These velocities are related by the velocity-addition formula, Equation 28.8. The velocity of the ions relative to
the alien spacecraft is:
(c) The aliens see the laser light (photons) moving with respect to the cruiser at a velocity
(d)The aliens see the ions moving away from the cruiser at a velocity
*40.
**41.
Two identical spaceships are under construction. The constructed length of each spaceship is 1.50 km. After
being launched, spaceship A moves away from earth at a constant velocity (speed is 0.850c) with respect to the earth.
Spaceship B follows in the same direction at a different constant velocity (speed is 0.500c) with respect to the earth.
Determine the length that a passenger on one spaceship measures for the other spaceship.
Two atomic particles approach each other in a head-on collision. Each particle has a mass of
. The
speed of each particle is
when measured by an observer standing in the laboratory.
(a)What is the speed of one particle as seen by the other particle?
Answer:
(b)Determine the magnitude of the relativistic momentum of one particle, as it would be observed by the other.
Answer:
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Problems
Section 29.3 Photons and the Photoelectric Effect
1.The dissociation energy of a molecule is the energy required to break the molecule apart into its separate atoms. The
dissociation energy for the cyanogen molecule is
photon. Determine the
(a)wavelength and
Answer:
. Suppose that this energy is provided by a single
(b)frequency of the photon.
Answer:
(c)In what region of the electromagnetic spectrum (see Figure 24.9) does this photon lie?
Answer:
ultraviolet region
2.An AM radio station broadcasts an electromagnetic wave with a frequency of 665 kHz, whereas an FM station
broadcasts an electromagnetic wave with a frequency of 91.9 MHz. How many AM photons are needed to have a total
energy equal to that of one FM photon?
3.
Ultraviolet light with a frequency of
strikes a metal surface and ejects electrons that have a
maximum kinetic energy of 6.1 eV. What is the work function (in eV) of the metal?
Answer:
6.3 eV
REASONING According to Equation 29.3, the work function
is related to the photon energy
and the
maximum kinetic energy
by
. This expression can be used to find the work function of
the metal.
SOLUTION
is 6.1 eV. The photon energy (in eV) is, according to Equation 29.2,
The work function is, therefore,
4.
Light is shining perpendicularly on the surface of the earth with an intensity of
. Assuming that all
the photons in the light have the same wavelength (in vacuum) of 730 nm, determine the number of photons per
second per square meter that reach the earth.
Ultraviolet light is responsible for sun tanning. Find the wavelength (in nm) of an ultraviolet photon whose
5.
energy is
Answer:
310 nm
.
REASONING The energy of the photon is related to its frequency by Equation 29.2,
. Equation 16.1,
, relates the frequency and the wavelength for any wave.
SOLUTION Combining Equations 29.2 and 16.1, and noting that the speed of a photon is c, the speed of light in a
vacuum, we have
6.The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 485
nm. What is the work function
of this metal? Express your answer in electron volts.
7.Radiation of a certain wavelength causes electrons with a maximum kinetic energy of 0.68 eV to be ejected from a
metal whose work function is 2.75 eV. What will be the maximum kinetic energy (in eV) with which this same
radiation ejects electrons from another metal whose work function is 2.17 eV?
Answer:
1.26 eV
8.
Multiple-Concept Example 3 reviews the concepts necessary to solve this problem. Radiation with a wavelength
of 238 nm shines on a metal surface and ejects electrons that have a maximum speed of
. Which one
of the following metals is it, the values in parentheses being the work functions: potassium (2.24 eV), calcium (2.71
eV), uranium (3.63 eV), aluminum (4.08 eV), or gold (4.82 eV)?
*9.
An owl has good night vision because its eyes can detect a light intensity as small as
. What is the minimum number of photons per second that an owl eye can detect if its pupil has a diameter of 8.5 mm
and the light has a wavelength of 510 nm?
Answer:
73 photons/s
REASONING AND SOLUTION The number of photons per second, N, entering the owl's eye is
, where S is the intensity of the beam, A is the area of the owl's pupil, and E is the energy of a single
photon. Assuming that the owl's pupil is circular,
Combining Equations 29.2 and 16.1, we have
, where d is the diameter of the owl's pupil.
. Therefore,
*10.
A proton is located at a distance of 0.420 m from a point charge of
. The repulsive electric force
moves the proton until it is at a distance of 1.58 m from the charge. Suppose that the electric potential energy lost by
the system were carried off by a photon. What would be its wavelength?
*11.When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum
kinetic energy of
. Determine the wavelength (in nm) of light that should be used to double the
maximum kinetic energy of the electrons ejected from this surface.
Answer:
162 nm
*12.
A glass plate has a mass of 0.50 kg and a specific heat capacity of
. The wavelength of
infrared light is
, while the wavelength of blue light is
. Find the number of infrared
photons and the number of blue photons needed to raise the temperature of the glass plate by
, assuming that
all the photons are absorbed by the glass.
**13.
A laser emits
photons per second in a beam of light that has a diameter of 2.00 mm and a
wavelength of 514.5 nm. Determine
(a)the average electric field strength and
Answer:
7760 N/C
REASONING AND SOLUTION
(a)
According to Equation 24.5b, the electric field can be found from
the beam is
where N is the number of photons per second emitted. Then,
(b)According to Equation 24.3, the average magnetic field is
. The intensity S of
(b)the average magnetic field strength for the electromagnetic wave that constitutes the beam.
Answer:
REASONING AND SOLUTION
(a)
According to Equation 24.5b, the electric field can be found from
the beam is
. The intensity S of
where N is the number of photons per second emitted. Then,
(b)According to Equation 24.3, the average magnetic field is
REASONING AND SOLUTION
(a)
According to Equation 24.5b, the electric field can be found from
is
. The intensity S of the beam
where N is the number of photons per second emitted. Then,
(b)According to Equation 24.3, the average magnetic field is
**14.
(a)How many photons
water at
?
(b)On the average, how many
must be absorbed to melt a 2.0-kg block of ice at
molecules does one photon convert from the ice phase to the water phase?
Section 29.4 The Momentum of a Photon and the Compton Effect
15.
into
A light source emits a beam of photons, each of which has a momentum of
(a)What is the frequency of the photons?
Answer:
.
(b)To what region of the electromagnetic spectrum do the photons belong? Consult Figure 24.9 if necessary.
Answer:
infrared region
16.
A photon of red light
and a Ping-Pong ball (
) have the same
momentum. At what speed is the ball moving?
In a Compton scattering experiment, the incident X-rays have a wavelength of 0.2685 nm, and the scattered X17.
rays have a wavelength of 0.2703 nm. Through what angle in Figure 29.10 are the X-rays scattered?
Answer:
REASONING The angle through which the X-rays are scattered is related to the difference between the
wavelength of the scattered X-rays and the wavelength of the incident X-rays by Equation 29.7 as
where h is Planck's constant, m is the mass of the electron, and c is the speed of light in a vacuum. We can use this
relation directly to find the angle, since all the other variables are known.
SOLUTION Solving Equation 29.7 for the angle , we obtain
18.
A sample is bombarded by incident X-rays, and free electrons in the sample scatter some of the X-rays at an
angle of
with respect to the incident X-rays (see Figure 29.10). The scattered X-rays have a momentum
whose magnitude is
. Determine the wavelength (in nm) of the incident X-rays. (For
accuracy, use
,
, and
for the mass of an
electron.)
An incident X-ray photon of wavelength 0.2750 nm is scattered from an electron that is initially at rest. The
19.
photon is scattered at an angle of
in Figure 29.10 and has a wavelength of 0.2825 nm. Use the
conservation of linear momentum to find the momentum gained by the electron.
Answer:
20.
In the Compton effect, momentum conservation applies, so the total momentum of the photon and the electron is
the same before and after the scattering occurs. Suppose that in Figure 29.10 the incident photon moves in the
direction and the scattered photon emerges at an angle of
, which is in the
direction. The incident
photon has a wavelength of
. Find the x and y components of the momentum of the scattered
electron.
*21.
What is the maximum amount by which the wavelength of an incident photon could change when it undergoes
Compton scattering from a nitrogen molecule
Answer:
?
REASONING The change in wavelength that occurs during Compton scattering is given by Equation 29.7:
is the maximum change in the wavelength, and to calculate it we need a value for the mass m of a
nitrogen molecule. This value can be obtained from the mass per mole M of nitrogen
and Avogadro's number
, according to
(see Section 14.1).
SOLUTION Using a value of
, we obtain the following result for the maximum change in the
wavelength:
*22.
Figure 29.10 shows the setup for measuring the Compton effect. With a fixed incident wavelength, a wavelength
of
is measured for a scattering angle of
angle of
*23.
, whereas a wavelength of
. Find the difference in wavelengths,
is measured for a scattering
.
A photon of wavelength 0.45000 nm strikes a free electron that is initially at rest. The photon is scattered
straight backward. What is the speed of the recoil electron after the collision?
Answer:
**24.An X-ray photon is scattered at an angle of
scattering, the electron has a speed of
from an electron that is initially at rest. After
. Find the wavelength of the incident X-ray photon.
Section
29.5 The De
Broglie
Wavelength
and the
Wave
Nature of
Matter
25.
A bacterium
bacterium?
Answer:
in the blood is moving at 0.33 m/s. What is the de Broglie wavelength of this
26.What are
(a)the wavelength of a 5.0-eV photon and
(b)the de Broglie wavelength of a 5.0-eV electron?
As Section 17.3 discusses, sound waves diffract, or bend, around the edges of a doorway. Larger wavelengths
27.
diffract more than smaller wavelengths.
(a)The speed of sound is 343 m/s. With what speed would a 55.0-kg person have to move through a doorway to
diffract to the same extent as a 128-Hz bass tone?
Answer:
REASONING In order for the person to diffract to the same extent as the sound wave, the de
Broglie wavelength of the person must be equal to the wavelength of the sound wave.
SOLUTION
(a) Since the wavelengths are equal, we have that
Solving for
, and using the relation
(Equation 16.1), we have
(b)At the speed calculated in part (a), the time required for the person to move a distance of one meter is
(b)At the speed calculated in part (a), how long (in years) would it take the person to move a distance of one
meter?
Answer:
REASONING In order for the person to diffract to the same extent as the sound wave, the de
Broglie wavelength of the person must be equal to the wavelength of the sound wave.
SOLUTION
(a) Since the wavelengths are equal, we have that
Solving for
, and using the relation
(Equation 16.1), we have
(b)At the speed calculated in part (a), the time required for the person to move a distance of one meter is
REASONING In order for the person to diffract to the same extent as the sound wave, the de Broglie
wavelength of the person must be equal to the wavelength of the sound wave.
SOLUTION
(a) Since the wavelengths are equal, we have that
Solving for
, and using the relation
(Equation 16.1), we have
(b)At the speed calculated in part (a), the time required for the person to move a distance of one meter is
28.An electron and a proton have the same speed. Ignore relativistic effects and determine the ratio
their de Broglie wavelengths.
29.
Recall from Section 14.3 that the average kinetic energy of an atom in a monatomic ideal gas is given by
where
helium atom (
Answer:
30.
and T is the Kelvin temperature of the gas. Determine the de Broglie wavelength of a
) that has the average kinetic energy at room temperature (293 K).
In a Young's double-slit experiment that uses electrons, the angle that locates the first-order bright fringes is
when the magnitude of the electron momentum is
the same double slit, what momentum magnitude
locates the first-order bright fringes?
*31.
of
. With
is necessary so that an angle of
A particle has a de Broglie wavelength of
. Then its kinetic energy doubles. What is the particle's
new de Broglie wavelength, assuming that relativistic effects can be ignored?
Answer:
REASONING The de Broglie wavelength is related to Planck's constant h and the magnitude p of the
particle's momentum. The magnitude of the momentum can be related to the particle's kinetic energy. Thus, using the
given wavelength and the fact that the kinetic energy doubles, we will be able to obtain the new wavelength.
SOLUTION The de Broglie wavelength is
(29.8)
The kinetic energy and the magnitude of the momentum are
(6.2)
(7.2)
where m and v are the mass and speed of the particle. Substituting Equation 7.2 into Equation 6.2, we can relate the
kinetic energy and momentum as follows:
Substituting this result for p into Equation 29.8 gives
Applying this expression for the final and initial wavelengths
Dividing the two equations and rearranging reveals that
and
, we obtain
Using the given value for
*32.
*33.
and the fact that
, we find
From a cliff that is 9.5 m above a lake, a young woman
jumps from rest, straight down into the
water. At the instant she strikes the water, what is her de Broglie wavelength?
The width of the central bright fringe in a diffraction pattern on a screen is identical when either electrons or
red light
pass through a single slit. The distance between the screen and the slit is
the same in each case and is large compared to the slit width. How fast are the electrons moving?
Answer:
*34.
*35.
Particle A is at rest, and particle B collides head-on with it. The collision is completely inelastic, so the two
particles stick together after the collision and move off with a common velocity. The masses of the particles are
different, and no external forces act on them. The de Broglie wavelength of particle B before the collision is
. What is the de Broglie wavelength of the object that moves off after the collision?
An electron, starting from rest, accelerates through a potential difference of 418 V. What is the final de Broglie
wavelength of the electron, assuming that its final speed is much less than the speed of light?
Answer:
REASONING The de Broglie wavelength of the electron is related to the magnitude p of its momentum by
(Equation 29.8), where h is Planck's constant. If the speed of the electron is much less than the speed of
light, the magnitude of the electron's momentum is given by
(Equation 7.2). Thus, the de Broglie wavelength
can be written as
.
When the electron is at rest, it has electric potential energy, but no kinetic energy. The electric potential energy EPE is
given by
(Equation 19.3), where e is the magnitude of the charge on the electron and V is the potential
difference. When the electron reaches its maximum speed, it has no potential energy, but its kinetic energy is
The conservation of energy states that the final total energy of the electron equals the initial total energy:
Solving this equation for the final speed gives
. Substituting this expression for
.
gives
.
SOLUTION After accelerating through the potential difference, the electron has a de Broglie wavelength of
**36.The kinetic energy of a particle is equal to the energy of a photon. The particle moves at 5.0% of the speed of light.
Find the ratio of the photon wavelength to the de Broglie wavelength of the particle.
Section 29.6 The Heisenberg Uncertainty Principle
37.
An object is moving along a straight line, and the uncertainty in its position is 2.5 m.
(a)Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's
velocity, assuming that the object is
Answer:
REASONING Suppose the object is moving along the
position is
. The minimum uncertainty
axis. The uncertainty in the object's
in the object's momentum is specified by the
Heisenberg uncertainty principle (Equation 29.10) in the form
. Since momentum
is mass m times velocity v, the uncertainty in the velocity
is related to the uncertainty in the momentum
by
.
SOLUTION
(a) Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows:
(b)For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by
(c) For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given
by
(b)a golf ball
Answer:
and
REASONING Suppose the object is moving along the
position is
. The minimum uncertainty
axis. The uncertainty in the object's
in the object's momentum is specified by the
Heisenberg uncertainty principle (Equation 29.10) in the form
. Since momentum
is mass m times velocity v, the uncertainty in the velocity
is related to the uncertainty in the momentum
by
.
SOLUTION
(a) Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows:
(b)For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by
(c) For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given
by
(c)an electron.
Answer:
REASONING Suppose the object is moving along the
position is
. The minimum uncertainty
axis. The uncertainty in the object's
in the object's momentum is specified by the
Heisenberg uncertainty principle (Equation 29.10) in the form
. Since momentum
is mass m times velocity v, the uncertainty in the velocity
is related to the uncertainty in the momentum
by
.
SOLUTION
(a) Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows:
(b)For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by
(c) For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given
by
REASONING Suppose the object is moving along the
. The minimum uncertainty
axis. The uncertainty in the object's position is
in the object's momentum is specified by the Heisenberg uncertainty
principle (Equation 29.10) in the form
. Since momentum is mass m times velocity v, the
uncertainty in the velocity
is related to the uncertainty in the momentum by
.
SOLUTION
(a) Using the uncertainty principle, we find the minimum uncertainty in the momentum as follows:
(b)For a golf ball this uncertainty in momentum corresponds to an uncertainty in velocity that is given by
(c) For an electron this uncertainty in momentum corresponds to an uncertainty in velocity that is given by
38.
A proton is confined to a nucleus that has a diameter of
. If this distance is considered to be the
uncertainty in the position of the proton, what is the minimum uncertainty in its momentum?
39.
In the lungs there are tiny sacs of air, which are called alveoli. An oxygen molecule (
)
is trapped within a sac, and the uncertainty in its position is 0.12 mm. What is the minimum uncertainty in the speed
of this oxygen molecule?
Answer:
REASONING The Heisenberg uncertainty principle specifies the relationship between the uncertainty
in a
particle's position and the uncertainty
in the particle's linear momentum. This principle is stated as follows:
(Equation 29.10), where
is Planck's constant. Since we seek the
minimum uncertainty in the speed (which is proportional to the momentum) of the oxygen molecule, we will use the
equals sign in Equation 29.10 and omit the greater-than symbol
uncertainty principle becomes
. For our purposes in this problem, then, the
(1)
We can use Equation 1 to calculate the minimum uncertainty in the speed, because the magnitude
of the
momentum is related to the speed according to
(Equation 7.2), where m is the mass of the oxygen
molecule.
SOLUTION Substituting
(Equation 7.2) into Equation 1, we obtain
(2)
Solving Equation 2 for the uncertainty
40.
*41.
in the speed, we find that
Particles pass through a single slit of width 0.200 mm (see Figure 29.14). The de Broglie wavelength of each
particle is 633 nm. After the particles pass through the slit, they spread out over a range of angles. Assume that the
uncertainty in the position of the particles is one-half the width of the slit, and use the Heisenberg uncertainty
principle to determine the minimum range of angles.
The minimum uncertainty
in the position y of a particle is equal to its de Broglie wavelength. Determine the
minimum uncertainty in the speed of the particle, where this minimum uncertainty
is expressed as a percentage of
the particle's speed
Answer:
8.0%
*42.
Assume that relativistic effects can be ignored.
A subatomic particle created in an experiment exists in a certain state for a time of
before
decaying into other particles. Apply both the Heisenberg uncertainty principle and the equivalence of energy and mass
(see Section 28.6) to determine the minimum uncertainty involved in measuring the mass of this short-lived particle.
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Problems
Section 30.1 Rutherford Scattering and the
Nuclear Atom
1.
The nucleus of the hydrogen atom has a radius of about
about
. The electron is normally at a distance of
from the nucleus. Assuming that the hydrogen atom is a sphere with a radius of
,
find
(a)the volume of the atom,
Answer:
REASONING Assuming that the hydrogen atom is a sphere of radius
given by
. Similarly, if the radius of the nucleus is
, the volume
, its volume
is
is given by
.
SOLUTION
(a)According to the given data, the nuclear dimensions are much smaller than the orbital radius of the
electron; therefore, we can treat the nucleus as a point about which the electron orbits. The electron is
normally at a distance of about
of radius
from the nucleus, so we can treat the atom as a sphere
. The volume of the atom is
(b Similarly, since the nucleus has a radius of approximately
)
, its volume is
(c) The percentage of the atomic volume occupied by the nucleus is
(b)the volume of the nucleus, and
Answer:
REASONING Assuming that the hydrogen atom is a sphere of radius
given by
. Similarly, if the radius of the nucleus is
, the volume
, its volume
is
is given by
.
SOLUTION
(a)According to the given data, the nuclear dimensions are much smaller than the orbital radius of the
electron; therefore, we can treat the nucleus as a point about which the electron orbits. The electron is
normally at a distance of about
of radius
from the nucleus, so we can treat the atom as a sphere
. The volume of the atom is
(b Similarly, since the nucleus has a radius of approximately
)
(c) The percentage of the atomic volume occupied by the nucleus is
, its volume is
(c)the percentage of the volume of the atom that is occupied by the nucleus.
Answer:
REASONING Assuming that the hydrogen atom is a sphere of radius
given by
. Similarly, if the radius of the nucleus is
, its volume
, the volume
is
is given by
.
SOLUTION
(a)According to the given data, the nuclear dimensions are much smaller than the orbital radius of the
electron; therefore, we can treat the nucleus as a point about which the electron orbits. The electron is
normally at a distance of about
of radius
from the nucleus, so we can treat the atom as a sphere
. The volume of the atom is
(b Similarly, since the nucleus has a radius of approximately
)
, its volume is
(c) The percentage of the atomic volume occupied by the nucleus is
REASONING Assuming that the hydrogen atom is a sphere of radius
, its volume
is given by
. Similarly, if the radius of the nucleus is
, the volume
is given by
.
SOLUTION
(a) According to the given data, the nuclear dimensions are much smaller than the orbital radius of the electron;
therefore, we can treat the nucleus as a point about which the electron orbits. The electron is normally at a
distance of about
from the nucleus, so we can treat the atom as a sphere of radius
. The volume of the atom is
(b)Similarly, since the nucleus has a radius of approximately
(c) The percentage of the atomic volume occupied by the nucleus is
, its volume is
2.
The nucleus of a hydrogen atom is a single proton, which has a radius of about
. The single electron in
a hydrogen atom normally orbits the nucleus at a distance of
. What is the ratio of the density of the
hydrogen nucleus to the density of the complete hydrogen atom?
3.Review Conceptual Example 1 and use the information therein as an aid in working this problem. Suppose that you're
building a scale model of the hydrogen atom, and the nucleus is represented by a ball that has a radius of 3.2 cm
(somewhat smaller than a baseball). How many miles away
Answer:
2 mi
4.
should the electron be placed?
In a Rutherford scattering experiment a target nucleus has a diameter of
. The incoming
particle has a mass of
. What is the kinetic energy of an particle that has a de Broglie wavelength
equal to the diameter of the target nucleus? Ignore relativistic effects.
*5.
There are Z protons in the nucleus of an atom, where Z is the atomic number of the element. An particle carries
a charge of
. In a scattering experiment, an particle, heading directly toward a nucleus in a metal foil, will come
to a halt when all the particle's kinetic energy is converted to electric potential energy. In such a situation, how close
will an particle with a kinetic energy of
Answer:
come to a gold nucleus
?
REASONING The distance of closest approach can be obtained by setting the kinetic energy KE of the
particle equal to the electric potential energy EPE of the
particle. According to Equation 19.3, we have that
, where
is the charge on the particle and V is the electric potential created by a gold nucleus.
According to Equation 19.6, the electric potential of the gold nucleus
is
where r is the distance between the a particle and the gold nucleus. Therefore, we have that
,
(1)
In this expression, we note that
and
SOLUTION Solving Equation 1 for the distance r we obtain
.
*6.
The nucleus of a copper atom contains 29 protons and has a radius of
. How much work (in
electron volts) is done by the electric force as a proton is brought from infinity, where it is at rest, to the “surface” of a
copper nucleus?
Section 30.2 Line Spectra, Section 30.3 The Bohr Model of the Hydrogen Atom
7.
For a doubly ionized lithium atom
, what is the principal quantum number of the state in which
the electron has the same total energy as a ground-state electron has in the hydrogen atom?
Answer:
REASONING According to the Bohr model, the energy
Equation 30.13:
electron in a doubly ionized lithium atom
(in eV) of the electron in an orbit is given by
. In order to find the principal quantum number of the state in which the
has the same total energy as a ground state electron in a hydrogen
atom, we equate the right hand sides of Equation 30.13 for the hydrogen atom and the lithium ion. This gives
This expression can be evaluated to find the desired principal quantum number.
SOLUTION For hydrogen,
, and
for the ground state. For lithium
,
. Therefore,
8.
A singly ionized helium atom
has only one electron in orbit about the nucleus. What is the radius of the ion
when it is in the second excited state?
orbit of a triply ionized beryllium atom
9.Using the Bohr model, determine the ratio of the energy of the
to the energy of the
orbit of a hydrogen atom (H).
Answer:
16
The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional 2.86 eV of
10.
energy. What is the quantum number n of the state into which the electron moves?
Find the energy (in joules) of the photon that is emitted when the electron in a hydrogen atom undergoes a
11.
transition from the
energy level to produce a line in the Paschen series.
Answer:
REASONING According to Equation 30.14, the wavelength emitted by the hydrogen atom when it makes a
transition from the level with to the level with is given by
where
and
for hydrogen. Once the wavelength for the particular
transition in question is determined, Equation 29.2
can be used to find the energy of the emitted
photon.
SOLUTION In the Paschen series,
. Using the above expression with
,
and
, we find that
The photon energy is
12.
excited state?
(a)What is the ionization energy of a hydrogen atom that is in the
excited state to the
(b)For a hydrogen atom, determine the ratio of the ionization energy for the
ionization energy for the ground state.
excited state. The atom
13.A hydrogen atom is in the ground state. It absorbs energy and makes a transition to the
returns to the ground state by emitting two photons. What are their wavelengths?
Answer:
and
14.In the hydrogen atom, what is the total energy (in electron volts) of an electron that is in an orbit that has a radius of
?
15.
Consider the Bohr energy expression (Equation 30.13) as it applies to singly ionized helium
and doubly
ionized lithium
. This expression predicts equal electron energies for these two species for certain
values of the quantum number n (the quantum number is different for each species). For quantum numbers less than or
equal to 9, what are the lowest three energies (in electron volts) for which the helium energy level is equal to the
lithium energy level?
Answer:
*16.
A sodium atom
contains 11 protons in its nucleus. Strictly speaking, the Bohr model does not apply,
because the neutral atom contains 11 electrons instead of a single electron. However, we can apply the model to the
outermost electron as an approximation, provided that we use an effective value
rather than 11 for the
number of protons in the nucleus.
(a)The ionization energy for the outermost electron in a sodium atom is 5.1 eV. Use the Bohr model with
to calculate a value for
.
(b)Using
and
, determine the corresponding two values for the radius of the outermost
Bohr orbit.
*17.
A wavelength of 410.2 nm is emitted by the hydrogen atoms in a high-voltage discharge tube. What are the
initial and final values of the quantum number n for the energy level transition that produces this wavelength?
Answer:
and
REASONING A wavelength of 410.2 nm is emitted by the hydrogen atoms in a high-voltage discharge tube.
This transition lies in the visible region (380–750 nm) of the hydrogen spectrum. Thus, we can conclude that the
transition is in the Balmer series and, therefore, that
. The value of can be found using Equation 30.14,
according to which the Balmer series transitions are given by
This expression may be solved for for the energy transition that produces the given wavelength.
SOLUTION Solving for , we find that
Therefore, the initial and final states are identified by
*18.
.
A hydrogen atom emits a photon that has momentum with a magnitude of
This photon is emitted because the electron in the atom falls from a higher energy level into the
.
level. What is
the quantum number of the level from which the electron falls? Use a value of
for Planck's
constant.
*19.
For atomic hydrogen, the Paschen series of lines occurs when
, whereas the Brackett series occurs when
in Equation 30.14. Using this equation, show that the ranges of wavelengths in these two series overlap.
Answer:
The answer is a proof.
REASONING For either series of lines, the wavelengths can be obtained from
(Equation 30.14), where
nucleus,
, and
SOLUTION For the Paschen series,
is the number of protons in the hydrogen
.
. The range of wavelengths occurs for values of
to
.
Using Equation 30.14, we find that the shortest wavelength occurs
The longest wavelength in the Paschen series occurs for
and is
and is
For the Brackett series,
. The range of wavelengths occurs for values of
30.14, we find that the shortest wavelength occurs for
and is
The longest wavelength in the Brackett series occurs for
to
. Using Equation
and is
Since the longest wavelength in the Paschen series falls within the Brackett series, the wavelengths of the two series
overlap.
*20.
Doubly ionized lithium
and triply ionized beryllium
each emit a line spectrum. For
a certain series of lines in the lithium spectrum, the shortest wavelength is 40.5 nm. For the same series of lines in the
beryllium spectrum, what is the shortest wavelength?
**21.
Bohr orbit, in terms of Z, n, and the
(a)Derive an expression for the speed of the electron in the
constants k, e, and h. For the hydrogen atom, determine the speed in
Answer:
orbit and
(b)the
Answer:
orbit.
(c)the
Answer:
(d)Generally, when speeds are less than one-tenth the speed of light, the effects of special relativity can
be ignored. Are the speeds found in (b) and (c) consistent with ignoring relativistic effects in the
Bohr model?
Answer:
Yes.
**22.In the Bohr model of hydrogen, the electron moves in a circular orbit around the nucleus. Determine the
angular speed of the electron, in revolutions per second, when it is in
(a)the ground state and
(b)the first excited state.
Section
30.5 The
Quantum
Mechanical
Picture of
the
Hydrogen
Atom
23.A hydrogen atom is in its second excited state. Determine, according to quantum mechanics,
(a)the total energy (in eV) of the atom,
Answer:
(b)the magnitude of the maximum angular momentum the electron can have in this state, and
Answer:
(c)the maximum value that the z component
Answer:
24.
of the angular momentum can have.
The table lists quantum numbers for five states of the hydrogen atom. Which (if any) of them are not possible?
For those that are not possible, explain why.
n
(a)
3
3
(b)
2
1
(c)
4
2
(d)
5
(e)
4
0
3
2
0
0
. What is the smallest possible value (the
25.The orbital quantum number for the electron in a hydrogen atom is
most negative) for the total energy of this electron? Give your answer in electron volts.
Answer:
26.
It is known that the possible values for the magnetic quantum number
are
,
,
,
, ,
,
,
, and
. Determine the orbital quantum number and the smallest possible value of the principle quantum
number n.
27.
The maximum value for the magnetic quantum number in state A is
, while in state B it is
. What
is the ratio
of the magnitudes of the orbital angular momenta of an electron in these two states?
Answer:
1.732
REASONING The maximum value for the magnetic quantum number is
; thus, in state A,
,
while in state B,
. According to the quantum mechanical theory of angular momentum, the magnitude of the
orbital angular momentum for a state of given is
(Equation 30.15). This expression can be
used to form the ratio
of the magnitudes of the orbital angular momenta for the two states.
SOLUTION Using Equation 30.15, we find that
*28.
*29.
The electron in a certain hydrogen atom has an angular momentum of
. What is the largest
possible magnitude for the z component of the angular momentum of this electron? For accuracy, use
.
For an electron in a hydrogen atom, the z component of the angular momentum has a maximum value of
. Find the three smallest possible values (the most negative) for the total energy (in electron
volts) that this atom could have.
Answer:
REASONING The total energy
for a hydrogen atom in the quantum mechanical picture is the same as in
the Bohr model and is given by Equation 30.13:
(30.13)
Thus, we need to determine values for the principal quantum number n if we are to calculate the three smallest
possible values for E. Since the maximum value of the orbital quantum number is
, we can obtain a minimum
value for n as
. But how to obtain ? It can be obtained, because the problem statement gives the
maximum value of , the z component of the angular momentum. According to Equation 30.16, is
(30.16)
where
is the magnetic quantum number and h is Planck's constant. For a given value of the allowed values for
are as follows:
. Thus, the maximum value of
is , and we can use
Equation 30.16 to calculate the maximum value of
from the maximum value given for .
SOLUTION Solving Equation 30.16 for
gives
**30.
As explained in the REASONING, this maximum value for
n is
indicates that
. Therefore, a minimum value for
This means that the three energies we seek correspond to
them to be
,
. Using Equation 30.13, we find
, and
An electron is in the
state. What is the smallest possible value for the angle between the z component of
the orbital angular momentum and the orbital angular momentum?
Section 30.6 The Pauli Exclusion Principle and the Periodic Table of the Elements
31.
Two of the three electrons in a lithium atom have quantum numbers of
,
,
,
(a)its ground state and
Answer:
,
,
,
What quantum numbers can the third electron have if the atom is in
n
2
0
0
2
0
0
1
1
1/2
(b)its first excited state?
Answer:
n
2
1/2
and
2
1
1
2
1
0
2
1
0
2
1
2
1
1/2
32.Following the style used in Table 30.3, determine the electronic configuration of the ground state for yttrium Y
. Refer to Figure 30.16 to see the order in which the subshells fill.
shell holds 2 electrons,
33.Figure 30.16 was constructed using the Pauli exclusion principle and indicates that the
the
shell holds 8 electrons, and the
shell holds 18 electrons. These numbers can be obtained by adding
the numbers given in the figure for the subshells contained within a given shell. How many electrons can be put into
the
shell, which is only partly shown in the figure?
Answer:
50
34.
Which of the following subshell configurations are not allowed? For those that are not allowed, give the reason
why.
(a)
(b)
(c)
(d)
(e)
When an electron makes a transition between energy levels of an atom, there are no restrictions on the
initial and final values of the principal quantum number n. According to quantum mechanics, however, there is a
rule that restricts the initial and final values of the orbital quantum number . This rule is called a selection rule
and states that
. In other words, when an electron makes a transition between energy levels, the value
of can only increase or decrease by one. The value of may not remain the same nor may it increase or decrease
by more than one. According to this rule, which of the following energy level transitions are allowed?
(a)
Answer:
not allowed
35.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the
initial and final values of the orbital quantum number . The selection rule states that when an electron
makes a transition between energy levels, the value of may not remain the same or increase or
decrease by more than one. In other words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to
the 1s state
. Since the value of is the same in both states,
, and we can
conclude that this energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
the 1s state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(c) For the transition
, the electron makes a transition from the 4p state
the 2p state
. Since the value of is the same in both states,
to
, and we can
conclude that this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
the 2p state
. The value of changes so that
conclude that this energy level transition is
.
to
, and we can
(e) For the transition
the 3s state
, the electron makes a transition from the 3d state
. The value of changes so that
conclude that this energy level transition is
(b)
Answer:
allowed
to
, and we can
.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the
initial and final values of the orbital quantum number . The selection rule states that when an electron
makes a transition between energy levels, the value of may not remain the same or increase or
decrease by more than one. In other words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to
the 1s state
. Since the value of is the same in both states,
, and we can
conclude that this energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
the 1s state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(c) For the transition
, the electron makes a transition from the 4p state
the 2p state
. Since the value of is the same in both states,
to
, and we can
conclude that this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
the 2p state
. The value of changes so that
conclude that this energy level transition is
(e) For the transition
the 3s state
to
, and we can
.
, the electron makes a transition from the 3d state
. The value of changes so that
conclude that this energy level transition is
(c)
Answer:
not allowed
to
, and we can
.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the
initial and final values of the orbital quantum number . The selection rule states that when an electron
makes a transition between energy levels, the value of may not remain the same or increase or
decrease by more than one. In other words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to
the 1s state
. Since the value of is the same in both states,
, and we can
conclude that this energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
the 1s state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(c) For the transition
, the electron makes a transition from the 4p state
the 2p state
. Since the value of is the same in both states,
to
, and we can
conclude that this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
the 2p state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(e) For the transition
, the electron makes a transition from the 3d state
the 3s state
. The value of changes so that
conclude that this energy level transition is
(d)
Answer:
allowed
to
, and we can
.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the
initial and final values of the orbital quantum number . The selection rule states that when an electron
makes a transition between energy levels, the value of may not remain the same or increase or
decrease by more than one. In other words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to
the 1s state
. Since the value of is the same in both states,
, and we can
conclude that this energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
the 1s state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(c) For the transition
, the electron makes a transition from the 4p state
the 2p state
. Since the value of is the same in both states,
to
, and we can
conclude that this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
the 2p state
. The value of changes so that
to
, and we can
conclude that this energy level transition is
.
(e) For the transition
, the electron makes a transition from the 3d state
the 3s state
. The value of changes so that
conclude that this energy level transition is
(e)
Answer:
not allowed
to
, and we can
.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the
initial and final values of the orbital quantum number . The selection rule states that when an electron
makes a transition between energy levels, the value of may not remain the same or increase or
decrease by more than one. In other words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to
the 1s state
. Since the value of is the same in both states,
, and we can
conclude that this energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
the 1s state
. The value of changes so that
conclude that this energy level transition is
.
(c) For the transition
, the electron makes a transition from the 4p state
the 2p state
. Since the value of is the same in both states,
conclude that this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
to
, and we can
to
, and we can
to
the 2p state
. The value of changes so that
conclude that this energy level transition is
(e) For the transition
, and we can
.
, the electron makes a transition from the 3d state
the 3s state
. The value of changes so that
conclude that this energy level transition is
to
, and we can
.
REASONING In the theory of quantum mechanics, there is a selection rule that restricts the initial and
final values of the orbital quantum number . The selection rule states that when an electron makes a transition
between energy levels, the value of may not remain the same or increase or decrease by more than one. In other
words, the rule requires that
.
SOLUTION
(a) For the transition
, the electron makes a transition from the 2s state
to the 1s state
. Since the value of is the same in both states,
, and we can conclude that this
energy level transition is
.
(b)For the transition
, the electron makes a transition from the 2p state
state
. The value of changes so that
energy level transition is
(c) For the transition
state
to the 1s
, and we can conclude that this
.
, the electron makes a transition from the 4p state
. Since the value of is the same in both states,
, and we can conclude that
this energy level transition is
.
(d)For the transition
, the electron makes a transition from the 4s state
state
. The value of changes so that
. The value of changes so that
energy level transition is
*36.In the ground state, the outermost shell
to the 2p
, and we can conclude that this
energy level transition is
.
(e) For the transition
, the electron makes a transition from the 3d state
state
to the 2p
to the 3s
, and we can conclude that this
.
of helium (He) is filled with electrons, as is the outermost shell
of neon (Ne). The full outermost shells of these two elements distinguish them as the first two so-called
“noble gases.” Suppose that the spin quantum number
had three possible values, rather than two. If that were
the case, which elements would be
(a)the first and
(b)the second noble gases? Assume that the possible values for the other three quantum numbers are
unchanged, and that the Pauli exclusion principle still applies.
Section
30.7 XRays
37.
By using the Bohr model, decide which element is likely to emit a
X-ray with a wavelength of
.
Answer:
carbon
38.
39.
What is the minimum potential difference that must be applied to an X-ray tube to knock a K-shell electron
completely out of an atom in a copper
target? Use the Bohr model as needed.
An X-ray tube is being operated at a potential difference of 52.0 kV. What is the Bremsstrahlung wavelength
that corresponds to 35.0% of the kinetic energy with which an electron collides with the metal target in the tube?
Answer:
40.
In the X-ray spectrum of niobium
,a
peak is observed at a wavelength of
.
(a)Determine the magnitude of the difference between the observed wavelength of the
X-ray for niobium
and that predicted by the Bohr model.
(b)Express the magnitude of this difference as a percentage of the observed wavelength.
41.
When a certain element is bombarded with high-energy electrons,
X-rays that have an energy of 9890 eV are
emitted. Determine the atomic number Z of the element, and identify the element. Use the Bohr model as necessary.
Answer:
(germanium, Ge)
REASONING As discussed in text Example 11, the wavelengths of the electrons depend upon the quantity
, where Z is the atomic number of the target. The Bohr model predicts that the energies of the electrons are
given by
(Equation 30.13), where n can take on any integer value greater than zero.
However, the photons correspond to a transition to the
shell from the
shell, and during this transition
the nuclear charge is partly screened by the electron that remains in the
shell, as discussed in text Example 10.
We will replace Z with
in Equation 30.13 to account for this screening. Therefore, the energy of the electron's
initial and final states are calculated from
(1)
The energy E of the emitted X-ray is equal to the higher initial energy
:
of the electron minus the lower final energy
(2)
SOLUTION Substituting Equation 1 into Equation 2, we obtain
(3)
Solving Equation 3 for
and taking the square root of both sides yields
The prediction of the Bohr model is, then,
The closest integer to this result is
*42.
*43.
; the atomic number of
.
The Bohr model, although not strictly applicable, can be used to estimate the minimum energy
that an
incoming electron must have in an X-ray tube in order to knock a K-shell electron entirely out of an atom in the metal
target. The
X-ray wavelength of metal A is 2.0 times the
X-ray wavelength of metal B. What is the ratio of
for metal A to
for metal B?
An X-ray tube contains a silver
target. The high voltage in this tube is increased from zero. Using the
Bohr model, find the value of the voltage at which the
X-ray just appears in the X-ray spectrum.
Answer:
21 600 V
REASONING In the spectrum of X-rays produced by the tube, the cutoff wavelength
and the voltage V of
the tube are related according to Equation 30.17,
. Since the voltage is increased from zero until the
X-ray just appears in the spectrum, it follows that
we find that
In this expression we have replaced Z with
text.
SOLUTION The desired voltage is, then,
*44.
and
. Using Equation 30.14 for
,
, in order to account for shielding, as explained in Example 11 in the
Multiple-Concept Example 9 reviews the concepts that are important in this problem. An electron, traveling at
a speed of
, strikes the target of an X-ray tube. Upon impact, the electron decelerates to one-quarter
of its original speed, an X-ray photon being emitted in the process. What is the wavelength of the photon?
Section 30.8 The Laser
45.
A laser is used in eye surgery to weld a detached retina back into place. The wavelength of the laser beam
is 514 nm, and the power is 1.5 W. During surgery, the laser beam is turned on for 0.050 s. During this time, how
many photons are emitted by the laser?
Answer:
REASONING The number of photons emitted by the laser will be equal to the total energy carried in the
beam divided by the energy per photon.
SOLUTION The total energy carried in the beam is, from the definition of power,
The energy of a single photon is given by Equations 29.2 and 16.1 as
where we have used the fact that
. Therefore, the number of photons emitted by the laser is
46.
The ultraviolet excimer laser used in the PRK technique (see Section 30.9) has a wavelength of 193 nm. A
carbon dioxide laser produces a wavelength of
. What is the minimum number of photons that the
carbon dioxide laser must produce to deliver at least as much or more energy to a target as does a single photon from
the excimer laser?
47.
A pulsed laser emits light in a series of short pulses, each having a duration of 25.0 ms. The average power of
each pulse is 5.00 mW, and the wavelength of the light is 633 nm. Find the number of photons in each pulse.
Answer:
48.
The drawing shows three energy levels of a laser that are involved in the lasing action. These levels are analogous
to the levels in the Ne atoms of a He-Ne laser. The
level is a metastable level, and the
level is the ground state.
The difference between the energy levels of the laser is shown in the drawing.
(a)What energy (in eV per electron) must an external source provide to start the lasing action?
(b)What is the wavelength of the laser light?
(c)In what region of the electromagnetic spectrum (see Figure 24.9) does the laser light lie?
49.
A laser peripheral iridotomy is a procedure for treating an eye condition known as narrow-angle glaucoma, in
which pressure buildup in the eye can lead to loss of vision. A neodymium YAG laser
used in the procedure to punch a tiny hole in the peripheral iris, thereby relieving the pressure buildup. In one
application the laser delivers
deliver?
Answer:
*50.
is
of energy to the iris in creating the hole. How many photons does the laser
Fusion is the process by which the sun produces energy. One experimental technique for creating controlled
fusion utilizes a solid-state laser that emits a wavelength of 1060 nm and can produce a power of
pulse duration of
W for a
. In contrast, the helium/neon laser used in a bar-code scanner at the checkout counter
emits a wavelength of 633 nm and produces a power of about
. How long (in days) would the
helium/neon laser have to operate to produce the same number of photons that the solid-state laser produces in
?
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Problems
Section 31.1 Nuclear Structure,
Section 31.2 The Strong Nuclear Force and the Stability of the Nucleus
1.
For
find
(a)the net electrical charge of the nucleus,
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the
nucleus is
nucleus contains 82 protons, the net electrical charge of the
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass
of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(b)the number of neutrons,
Answer:
126
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the
nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass
of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(c)the number of nucleons,
Answer:
208
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the
nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass
of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(d)the approximate radius of the nucleus, and
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the
nucleus is
nucleus contains 82 protons, the net electrical charge of the
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass
of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
(e)the nuclear density.
Answer:
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
,
where A represents the total number of protons and neutrons (the nucleon number) and Z represents the
number of protons in the nucleus (the atomic number). The number of neutrons N is related to A and Z by
Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the
charge on a single proton. Since the
nucleus contains 82 protons, the net electrical charge of the
nucleus is
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be
found by multiplying the mass
of a single nucleon by the total number A of nucleons in the
nucleus. Treating the nucleus as a sphere of radius r, the nuclear density is
Therefore,
REASONING For an element whose chemical symbol is X, the symbol for the nucleus is
, where A
represents the total number of protons and neutrons (the nucleon number) and Z represents the number of protons in
the nucleus (the atomic number). The number of neutrons N is related to A and Z by Equation 31.1:
.
SOLUTION For the nucleus
, we have
and
.
(a) The net electrical charge of the nucleus is equal to the total number of protons multiplied by the charge on a
single proton. Since the
nucleus contains 82 protons, the net electrical charge of the
(b)The number of neutrons is
(c) By inspection, the number of nucleons is
nucleus is
.
.
(d)The approximate radius of the nucleus can be found from Equation 31.2, namely
(e) The nuclear density is the mass per unit volume of the nucleus. The total mass of the nucleus can be found by
multiplying the mass
of a single nucleon by the total number A of nucleons in the nucleus. Treating
the nucleus as a sphere of radius r, the nuclear density is
Therefore,
2.A nucleus contains 18 protons and 22 neutrons. What is the radius of this nucleus?
3.In each of the following cases, what element does the symbol X represent and how many neutrons are in the nucleus?
Use the periodic table on the inside of the back cover as needed.
(a)
Answer:
117 neutrons, platinum (Pt)
(b)
Answer:
16 neutrons, sulfur (S)
(c)
Answer:
34 neutrons, copper (Cu)
(d)
Answer:
6 neutrons, boron (B)
(e)
Answer:
145 neutrons, plutonium (Pu)
4.By what factor does the nucleon number of a nucleus have to increase in order for the nuclear radius to double?
In electrically neutral atoms, how many
5.
(a)
protons are in the uranium
Answer:
92 protons
nucleus,
REASONING
(a)
The number of protons in a given nucleus
is specified by its atomic number Z.
(b)
The number N of neutrons in a given nucleus
is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons):
(Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a)
The number of protons in the uranium
nucleus is
.
(b)
The number N of neutrons in the
nucleus is
.
(c)
The number of electrons that orbit the
nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or
.
(b)
neutrons are in the mercury
nucleus, and
Answer:
122 neutrons
REASONING
(a)
The number of protons in a given nucleus
is specified by its atomic number Z.
(b)
The number N of neutrons in a given nucleus
is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons):
(Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a)
The number of protons in the uranium
nucleus is
.
(b)
The number N of neutrons in the
nucleus is
.
(c)
The number of electrons that orbit the
nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or
(c)
electrons are in orbit about the niobium
Answer:
41 electrons
.
nucleus?
REASONING
(a)
The number of protons in a given nucleus
is specified by its atomic number Z.
(b)
The number N of neutrons in a given nucleus
is equal to the nucleon number A (the number of
protons and neutrons) minus the atomic number Z (the number of protons):
(Equation
31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to
the number of protons in the nucleus.
SOLUTION
(a)
The number of protons in the uranium
nucleus is
.
(b)
The number N of neutrons in the
nucleus is
.
(c)
The number of electrons that orbit the
nucleus in the neutral niobium atom is equal to the
number of protons in the nucleus, or
REASONING
.
(a)
The number of protons in a given nucleus
is specified by its atomic number Z.
(b)
The number N of neutrons in a given nucleus
is equal to the nucleon number A (the number of protons and
neutrons) minus the atomic number Z (the number of protons):
(Equation 31.1).
(c) In an electrically neutral niobium atom, the number of electrons in orbit about the nucleus is equal to the
number of protons in the nucleus.
SOLUTION
(a)
The number of protons in the uranium
nucleus is
.
(b)
The number N of neutrons in the
nucleus is
.
(c)
The number of electrons that orbit the
nucleus in the neutral niobium atom is equal to the number of
protons in the nucleus, or
.
6.
The largest stable nucleus has a nucleon number of 209, and the smallest has a nucleon number of 1. If each
nucleus is assumed to be a sphere, what is the ratio (largest/smallest) of the surface areas of these spheres?
*7.
The ratio
of the radius of an unknown nucleus
to the radius of a tritium nucleus
Both nuclei contain the same number of neutrons. Identify the unknown nucleus in the form
table on the inside of the back cover as needed.
Answer:
*8.
An unknown nucleus contains 70 neutrons and has twice the volume of the nickel
is
.
. Use the periodic
nucleus. Identify the
unknown nucleus in the form
. Use the periodic table on the inside of the back cover as needed.
**9.
Refer to Conceptual Example 1 for a discussion of nuclear densities. A neutron star is composed of
neutrons and has a density that is approximately the same as that of a nucleus. What is the radius of a neutron
star whose mass is 0.40 times the mass of the sun?
Answer:
REASONING According to Equation 31.2, the radius of a nucleus in meters is
, where A is the nucleon number. If we treat the neutron star as a uniform sphere, its
density (Equation 11.1) can be written as
Solving for the radius r, we obtain,
This expression can be used to find the radius of a neutron star of mass M and density .
SOLUTION As discussed in Conceptual Example 1, nuclear densities have the same approximate value in all
atoms. If we consider a uniform spherical nucleus, then the density of nuclear matter is approximately given by
The mass of the sun is
(see inside of the front cover of the text). Substituting values into the
expression for r determined above, we find
**10.
Suppose that you could pack neutrons
inside a tennis ball
in
the same way as neutrons and protons are packed together in the nucleus of an atom.
(a)Approximately how many neutrons would fit inside the tennis ball?
(b)A small object is placed 2.0 m from the center of the neutron-packed tennis ball, and the tennis ball
exerts a gravitational force on it. When the object is released, what is the magnitude of the acceleration
that it experiences? Ignore the gravitational force exerted on the object by the earth.
Section
31.3
The
Mass
Defect
of the
Nucleus
and
Nuclear
Binding
Energy
(Note: The atomic mass for hydrogen
is 1.007 825 u; this includes the mass of one electron.)
11.
Find the binding energy (in MeV) for lithium
.
Answer:
39.25 MeV
REASONING To obtain the binding energy, we will calculate the mass defect and then use the fact that 1 u is
equivalent to 931.5 MeV. The atomic mass given for
includes the 3 electrons in the neutral atom. Therefore, when
computing the mass defect, we must account for these electrons. We do so by using the atomic mass of 1.007 825 u
for the hydrogen atom
, which also includes the single electron, instead of the atomic mass of a proton.
SOLUTION Noting that the number of neutrons is
, we obtain the mass defect
as follows:
Since 1 u is equivalent to 931.5 MeV, the binding energy is
12.The binding energy of a nucleus is 225.0 MeV. What is the mass defect of the nucleus in atomic mass units?
13.Determine the mass defect (in atomic mass units) for
(a)
helium
, which has an atomic mass of 3.016 030 u, and
Answer:
0.008 285 u
(b)
the isotope of hydrogen known as tritium , which has an atomic mass of 3.016 050 u.
Answer:
0.009 105 u
(c)On the basis of your answers to parts (a) and (b), state which nucleus requires more energy to disassemble it
into its separate and stationary constituent nucleons. Give your reasoning.
Answer:
More energy must be supplied to
14.
than to helium
.
A 245-kg boulder is dropped into a mine shaft that is
deep. During the boulder's fall, the system
consisting of the earth and the boulder loses a certain amount of gravitational potential energy. It would take an equal
amount of energy to “free” the boulder from the shaft by raising it back to the top, so this can be considered the
system's binding energy.
(a)Determine the binding energy (in joules) of the earth–boulder system.
(b)How much mass does the earth–boulder system lose when the boulder falls to the bottom of the shaft?
15.
For lead
obtain
(a)the mass defect in atomic mass units,
Answer:
1.741 670 u
(b)the binding energy (in MeV), and
Answer:
1622 MeV
(c)the binding energy per nucleon (in MeV/nucleon).
Answer:
7.87 MeV/nucleon
*16.
(a)Energy is required to separate a nucleus into its constituent nucleons, as Figure 31.3 indicates; this energy is
the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single
nucleon to the remainder of the nucleus. For example, separating nitrogen
into nitrogen
neutron takes energy equal to the binding energy of the neutron, as shown below:
Find the energy (in MeV) that binds the neutron to the
and a
nucleus by considering the mass of
and the mass of
the mass of
.
(b)
Similarly, one can speak of the energy that binds a single proton to the
, as compared to
nucleus:
Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton
to the
nucleus. The atomic mass of carbon
is 13.003 355 u.
(c)Which nucleon is more tightly bound, the neutron or the proton?
*17.
Two isotopes of a certain element have binding energies that differ by 5.03 MeV. The isotope with the larger
binding energy contains one more neutron than the other isotope. Find the difference in atomic mass between the two
isotopes.
Answer:
1.003 27 u
REASONING Since we know the difference in binding energies for the two isotopes, we can determine the
corresponding mass defect. Also knowing that the isotope with the larger binding energy contains one more neutron
than the other isotope gives us enough information to calculate the atomic mass difference between the two isotopes.
SOLUTION The mass defect corresponding to a binding energy difference of 5.03 MeV is
Since the isotope with the larger binding energy has one more neutron
difference in atomic mass between the two isotopes is
than the other isotope, the
*18.
A copper penny has a mass of 3.0 g. Determine the energy (in MeV) that would be required to break all the
copper nuclei into their constituent protons and neutrons. Ignore the energy that binds the electrons to the nucleus and
the energy that binds one atom to another in the structure of the metal. For simplicity, assume that all the copper
nuclei are
.
Section 31.4 Radioactivity
19.
Write the
decay process for each of the following nuclei, being careful to include Z and A and the proper
chemical symbol for each daughter nucleus:
(a)
Answer:
REASONING As discussed in Section 31.4,
decay occurs when the nucleus emits a positron,
which has the same mass as an electron but carries a charge of
decay is
SOLUTION
(a)
Therefore, the
(b)
Similarly, the
(b)
decay process for
is
decay process for
is
instead of
. The general form for
.
.
Answer:
REASONING As discussed in Section 31.4,
decay occurs when the nucleus emits a positron,
which has the same mass as an electron but carries a charge of
decay is
SOLUTION
(a)
Therefore, the
(b)
Similarly, the
decay process for
is
decay process for
is
REASONING As discussed in Section 31.4,
the same mass as an electron but carries a charge of
SOLUTION
instead of
. The general form for
.
.
decay occurs when the nucleus emits a positron, which has
instead of
. The general form for
decay is
(a)
Therefore, the
(b)
Similarly, the
20.
decay process for
is
decay process for
is
.
.
Write the
decay process for carbon
, including the chemical symbols as well as the values of Z and A for the
parent and daughter nuclei and the
particle.
21.
Osmium
is converted into iridium
via
decay. What is the energy (in MeV) released in this process?
Answer:
0.313 MeV
REASONING The reaction and the atomic masses are:
When the
nucleus is converted into an iridium
nucleus, the number of orbital electrons remains the same,
so the resulting iridium atom is missing one orbital electron. However, the given mass includes all 77 electrons of a
neutral iridium atom. In effect, then, the value of 190.960 584 u for
already includes the mass of the
Since energy is released during the decay, the combined mass of the iridium
particle.
daughter nucleus and the
particle is less than the mass of the osmium
parent nucleus. The difference in mass is equivalent to the energy
released. To obtain the energy released in MeV, we will use the fact that 1 u is equivalent to 931.5 MeV.
SOLUTION The mass decrease that accompanies the
decay of osmium
is
. Since 1 u is equivalent to 931.5 MeV, the energy released is
22.
Find the energy that is released when a nucleus of lead
to become bismuth
undergoes
decay
.
23.
Find the energy (in MeV) released when decay converts radium
into radon
. The atomic mass of an particle is 4.002 603 u.
Answer:
4.87 MeV
24.
25.
Lead
is a stable daughter nucleus that can result from either an decay or a
decay. Write the decay
processes, including the chemical symbols and values for Z and A of the parent nuclei, for
(a)the decay and
(b)the
decay.
In the form
identify the daughter nucleus that results when
(a)
plutonium
undergoes decay,
Answer:
(b)
sodium
Answer:
undergoes
decay, and
(c)
nitrogen
Answer:
26.
undergoes
decay.
Multiple-Concept Example 7 reviews the concepts needed to solve this problem. When uranium
emits (among other things) a ray that has a wavelength of
ray.
decays, it
. Determine the energy (in MeV) of this
*27.
Polonium
undergoes decay. Assuming that all the released energy is
in the form of kinetic energy of the particle
nucleus (lead
Answer:
*28.
Radon
and ignoring the recoil of the daughter
, 205.974 440 u), find the speed of the particle. Ignore relativistic effects.
produces a daughter nucleus that is radioactive. The daughter, in turn, produces its own radioactive
daughter, and so on. This process continues until lead
is reached. What are the total number
and the total number
of
particles that are generated in this series of radioactive decays?
*29.
Review Conceptual Example 5 as background for this problem. The decay of uranium
of particles
produces thorium
. In Example 4, the energy released in this decay is determined to be 4.3 MeV.
Determine how much of this energy is carried away by the recoiling
daughter nucleus and how much by the
particle
. Assume that the energy of each particle is kinetic energy, and ignore the
small amount of energy carried away by the ray that is also emitted. In addition, ignore relativistic effects.
Answer:
0.072 MeV (thorium atom) and 4.2 MeV (alpha particle)
**30.
An isotope of beryllium
emits a ray and recoils with a speed of
Assuming that the beryllium nucleus is stationary to begin with, find the wavelength of the ray.
**31.
Find the energy (in MeV) released when
decay converts sodium
into neon
atomic mass for
includes the mass of 11 electrons, whereas the atomic mass for
of only 10 electrons.
Answer:
1.82 MeV
. Notice that the
includes the mass
REASONING Energy is released during the decay. To find the energy released, we determine how
much the mass has decreased because of the decay and then calculate the equivalent energy. The reaction and
masses are shown below:
SOLUTION The decrease in mass is
where the extra electron mass takes into account the fact that the atomic mass for sodium includes the mass of
11 electrons, whereas the atomic mass for neon includes the mass of only 10 electrons.
Since 1 u is equivalent to 931.5 MeV, the released energy is
.
Section
31.6
Radioactiv
e Decay
and
Activity
32.In 9.0 days the number of radioactive nuclei decreases to one-eighth the number present initially. What is the half-life
(in days) of the material?
The half-lives in two different samples, A and B, of radioactive nuclei are related according to
33.
. In a certain period the number of radioactive nuclei in sample A decreases to one-fourth the
number present initially. In this same period the number of radioactive nuclei in sample B decreases to a fraction f of
the number present initially. Find f.
Answer:
REASONING The basis of our solution is the fact that only one-half of the radioactive nuclei present initially
remain after a time equal to one half-life. After a time period that equals two half-lives, the number of nuclei
remaining is
of the initial number. After three half-lives, the number remaining is
of
the initial number, and so on. Thus, to determine the fraction of nuclei remaining after a given time period, we need
only to know how many half-lives that period contains.
SOLUTION For sample A, the number of nuclei remaining is
of the initial number. We can see, then,
that the time period involved is equal to two half lives of radioactive isotope A. But we know that
, so that this time period must be equal to four half-lives of radioactive isotope B. The fraction f
of the B nuclei that remain, therefore, is
34.
The
isotope of phosphorus has a half-life of 14.28 days. What is its decay constant in units of
?
35.
Strontium
has a half-life of 29.1 yr. It is chemically similar to calcium, enters the body through the food
chain, and collects in the bones. Consequently,
take for 99.9900% of the
Answer:
387 yr
36.
is a particularly serious health hazard. How long (in years) will it
released in a nuclear reactor accident to disappear?
Two radioactive waste products from nuclear reactors are strontium
and cesium
. These two species are present initially in a ratio of
. What is
the ratio
fifteen years later?
Suppose that the activity of a radioactive substance is initially 398 disintegrations/min and two days later it is
37.
285 disintegrations/min. What is the activity four days later still, or six days after the start? Give your answer in
disintegrations/min.
Answer:
146 disintegrations/min
REASONING We can find the decay constant from Equation 31.5,
by the decay constant , we have
. If we multiply both sides
where
is the initial activity and A is the activity after a time t. Once the decay constant is known, we can use the
same expression to determine the activity after a total of six days.
SOLUTION Solving the expression above for the decay constant , we have
Then the activity four days after the second day is
38.
Iodine
is used in diagnostic and therapeutic techniques in the treatment of thyroid disorders. This isotope
has a half-life of 8.04 days. What percentage of an initial sample of
39.
remains after 30.0 days?
The number of radioactive nuclei present at the start of an experiment is
twenty days later is
Answer:
8.00 days
. The number present
. What is the half-life (in days) of the nuclei?
REASONING AND SOLUTION According to Equation 31.5,
, the decay constant is
The half-life is, from Equation 31.6,
*40.One day, a cell phone company sends a text message to each of its 5800 subscribers, announcing that they have been
automatically enrolled as contestants in a promotional lottery modeled on nuclear decay. On the first day, 10% of the
5800 contestants are notified by text message that they have been randomly eliminated from the lottery. The other
90% of the contestants automatically advance to the next round. On each of the following days, 10% of the remaining
contestants are randomly eliminated, until fewer than 10 contestants remain. Determine
(a)the activity (number of contestants eliminated per day) on the second day of the lottery,
(b)
the decay constant (in
) of the lottery, and
(c)the half-life (in d) of the lottery.
*41.
A device used in radiation therapy for cancer contains 0.50 g of cobalt
(59.933 819 u). The half-life of
is 5.27 yr. Determine the activity of the radioactive material.
Answer:
*42.
A one-gram sample of radium
(
nuclei and undergoes decay to produce radon
mass of an particle is 4.002 603 u. The latent heat of fusion for water is
in 3.66 days, how many kilograms of ice could be melted at
?
,
) contains
. The atomic
. With the energy released
*43.
The isotope
of gold has a half-life of 2.69 days and is used in
cancer therapy. What mass (in grams) of this isotope is required to produce an activity of 315 Ci?
Answer:
*44.
Outside the nucleus, the neutron itself is radioactive and decays into a proton, an electron, and an antineutrino.
The half-life of a neutron
outside the nucleus is 10.4 min. On average, over what
distance (in meters) would a beam of 5.00-eV neutrons travel before the number of neutrons decreased to 75.0% of its
initial value?
*45.
Two radioactive nuclei A and B are present in equal numbers to begin with. Three days later, there are three
times as many A nuclei as there are B nuclei. The half-life of species B is 1.50 days. Find the half-life of species A.
Answer:
7.23 days
REASONING According to Equation 31.5, the number of nuclei remaining after a time t is
Using this expression, we find the ratio
as follows:
where we have used the fact that initially the numbers of the two types of nuclei are equal
the natural logarithm of both sides of the equation above shows that
SOLUTION Since
when
. Taking
days, it follows that
But we need to find the half-life of species B, so we use Equation 31.6, which indicates that
this expression for , the result for
becomes
Since
.
, the result above can be solved to show that
. With
.
Section 31.7 Radioactive Dating
46.
A sample has a
activity of 0.0061 Bq per gram of carbon.
(a)Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been
constant at a value of 0.23 Bq.
(b)Evidence suggests that the value of 0.23 Bq might have been as much as 40% larger. Repeat part (a), taking
into account this 40% increase.
Review Multiple-Concept Example 11 for help in approaching this problem. An archaeological specimen
47.
containing 9.2 g of carbon has an activity of 1.6 Bq. How old (in years) is the specimen?
Answer:
REASONING According to Equation 31.5, the number of nuclei remaining after a time t is
If we multiply both sides of this equation by the decay constant , we have
the activity A, we have
, where
is the activity at time
. Recognizing that
.
.
is
can be determined from the fact that
we know the mass of the specimen, and that the activity of one gram of carbon in a living organism is 0.23 Bq. The
decay constant can be determined from the value of 5730 yr for the half-life of
using Equation 31.6. With known
values for
and , the given activity of 1.6 Bq can be used to determine the age t of the specimen.
SOLUTION For
, the decay constant is
The activity at time
is
of the specimen can be determined from
. Since
and
, the age
Taking the natural logarithm of both sides leads to
Therefore, the age of the specimen is
48.
The half-life for the decay of uranium
is
. Determine the age (in years) of a rock specimen
that contains 60.0% of its original number of
atoms.
49.Review Conceptual Example 12 before starting to solve this problem. The number of unstable nuclei remaining after a
time
is N, and the number present initially is
. Find the ratio
for
(a)
,
Answer:
0.999
(b)
(
; use
, since otherwise the answer is out of the range of your
calculator), and
Answer:
(c)
. Verify that your answers are consistent with the reasoning in Conceptual
Example 12.
Answer:
0.755
50.
Multiple-Concept Example 11 reviews most of the concepts that are needed to solve this problem. Material found
with a mummy in the arid highlands of southern Peru has a
activity per gram of carbon that is 78.5% of the
activity present initially. How long ago (in years) did this individual die?
*51.
When any radioactive dating method is used, experimental error in the measurement of the sample's activity
leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity
of 0.100 Bq per gram of carbon is measured to within an accuracy of
. Find the age of the fossils and the
maximum error (in years) in the value obtained. Assume that there is no error in the 5730-year half-life of
the value of 0.23 Bq per gram of carbon in a living organism.
Answer:
REASONING According to Equation 31.5,
we have
nor in
. If we multiply both sides by the decay constant ,
where
is the initial activity and A is the activity after a time t. The decay constant is related to the half-life through
Equation 31.6:
. We can find the age of the fossils by solving for the time t. The maximum error
can be found by evaluating the limits of the accuracy as given in the problem statement.
SOLUTION The age of the fossils is
The maximum error can be found as follows. When there is an error of
, and we have
Similarly, when there is an error of
,
,
, and we have
The maximum error in the age of the fossils is
.
(a)A sample is being dated by the radiocarbon technique. If the sample were uncontaminated, its activity would
be 0.011 Bq per gram of carbon. Find the true age (in years) of the sample.
(b)Suppose the sample is contaminated, so that only 98.0% of its carbon is ancient carbon. The remaining 2.0%
**52.
is fresh carbon, in the sense that the
it contains has not had any time to decay. Assuming that the lab
technician is unaware of the contamination, what apparent age (in years) would be determined for the
sample?
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.
Problems
Section 32.1 Biological Effects of Ionizing
Radiation
1.
Neutrons
and particles have the same biologically equivalent dose. However, the absorbed dose
of the neutrons is six times the absorbed dose of the particles. What is the RBE for the particles?
Answer:
12
REASONING The biologically equivalent dose (in rems) is the product of the absorbed dose (in rads) and the
relative biological effectiveness (RBE), according to Equation 32.4. We can apply this equation to each type of
radiation. Since the biologically equivalent doses of the neutrons and particles are equal, we can solve for the
unknown RBE.
SOLUTION Applying Equation 32.4 to each type of particle and using the fact that the biological equivalent doses
are equal, we find that
Solving for
and noting that
2.What absorbed dose (in rads) of particles
, we have
causes as much biological damage as a 60-rad dose of
protons
?
3.
Over a full course of treatment, two different tumors are to receive the same absorbed dose of therapeutic
radiation. The smaller of the tumors
absorbs a total of 1.7 J of energy.
Determine
the
absorbed
dose,
in
Gy.
(a)
Answer:
14 Gy
(b)What is the total energy absorbed by the larger of the tumors
?
Answer:
2.1 J
4.Over a year's time, a person receives a biologically equivalent dose of 24 mrem (millirems) from cosmic rays, which
consist primarily of high-energy protons bombarding earth's atmosphere from space. The relative biological
effectiveness of protons is 10.
(a)What is the person's absorbed dose in rads?
(b)
The person absorbs
of energy from cosmic rays in a year. What is the person's mass?
5.
A beam of particles is directed at a 0.015-kg tumor. There are
particles per second reaching
the tumor, and the energy of each particle is 4.0 MeV. The RBE for the radiation is 14. Find the biologically
equivalent dose given to the tumor in 25 s.
Answer:
REASONING AND SOLUTION According to Equation 32.2, the absorbed dose (AD) is equal to the energy
absorbed by the tumor divided by its mass:
The biologically equivalent dose (BED) is equal to the product of the absorbed dose (AD) and the RBE (see Equation
32.4):
(32.4)
6.
A 75-kg person is exposed to 45 mrem of particles
absorbed?
. How much energy (in joules) has this person
7.
Multiple-Concept Example 1 discusses the concepts that are relevant to this problem. A person undergoing
radiation treatment for a cancerous growth receives an absorbed dose of 2.1 Gy. All the radiation is absorbed by the
growth. If the growth has a specific heat capacity of
Answer:
8.
The biologically equivalent dose for a typical chest X-ray is
21 kg, and it absorbs
this particular type of tissue?
, determine the rise in its temperature.
. The mass of the exposed tissue is
of energy. What is the relative biological effectiveness (RBE) for the radiation on
*9.
A 2.0-kg tumor is being irradiated by a radioactive source. The tumor receives an absorbed dose of 12 Gy in a
time of 850 s. Each disintegration of the radioactive source produces a particle that enters the tumor and delivers an
energy of 0.40 MeV. What is the activity
(see Section 31.6) of the radioactive source?
Answer:
*10.
Multiple-Concept Example 1 uses an approach similar to that needed in this problem, except here the
temperature remains constant while a phase change occurs. A sample of liquid water at
and 1 atm pressure
boils into steam at
because it is irradiated with a large dose of ionizing radiation. What is the absorbed dose of
the radiation in rads?
*11.
A beam of nuclei is used for cancer therapy. Each nucleus has an energy of 130 MeV, and the relative
biological effectiveness (RBE) of this type of radiation is 16. The beam is directed onto a 0.17-kg tumor, which
receives a biologically equivalent dose of 180 rem. How many nuclei are in the beam?
Answer:
REASONING The number of nuclei in the beam is equal to the energy absorbed by the tumor divided
by the energy per nucleus (130 MeV). According to Equation 32.2, the energy (in joules) absorbed by the tumor
is equal to the absorbed dose (expressed in grays) times the mass of the tumor. The absorbed dose (expressed in
rads) is equal to the biologically equivalent dose divided by the RBE of the radiation (see Equation 32.4). We
can use these concepts to determine the number of nuclei in the beam.
SOLUTION The number N of nuclei in the beam is equal to the energy E absorbed by the tumor divided by the
energy per nucleus. Since the energy absorbed is equal to the absorbed dose (in Gy) times the mass m (see
Equation 32.2) we have
We can express the absorbed dose in terms of rad units, rather than Gy units, by noting that
Therefore,
.
The number of nuclei can now be written as
We know from Equation 32.4 that the Absorbed dose (in rad) is equal to the biologically equivalent dose
divided by the RBE, so that
Section
32.2
Induced
Nuclear
Reactions
12.
What is the atomic number Z, the atomic mass number A, and the element X in the reaction
?
13.
Write the equation for the reaction
are produced by the reaction.
Answer:
. The notation “αn” means that an particle and a neutron
REASONING The reaction given in the problem statement is written in the shorthand form:
. The first and last symbols represent the initial and final nuclei, respectively. The symbols inside
the parentheses denote the incident particles or rays (left side of the comma) and the emitted particles or rays (right
side of the comma).
SOLUTION In the shorthand form of the reaction, we note that the designation an refers to an particle (which is a
helium nucleus
14.
) and a neutron
. Thus, the reaction is
For each of the nuclear reactions listed below, determine the unknown particle
inside of the back cover as needed.
(a)
. Use the periodic table on the
(b)
(c)
(d)
15.
A neutron causes
to change according to the reaction
(a)
Identify the unknown nucleus
for the element.
Answer:
, giving its atomic mass number A, its atomic number Z, and the symbol X
(b)
The
nucleus subsequently undergoes
decay, and its daughter does too. Identify the final nucleus,
giving its atomic mass number, atomic number, and name.
Answer:
16.Write the reactions below in the shorthand form discussed in the text.
(a)
(b)
(c)
17.
Complete the following nuclear reactions, assuming that the unknown quantity signified by the question mark is
a single entity:
(a)
Answer:
REASONING During each reaction, both the total electric charge of the nucleons and the total
number of nucleons are conserved. For each type of nucleus that participates in a reaction, the atomic mass
number A specifies the number of nucleons, and the atomic number Z specifies the number of protons. We
will use these conserved quantities to calculate the atomic mass number and atomic number for each of the
unknown entities, thereby identifying them.
SOLUTION
(a)
In the reaction described by
?, a neutron
induces a reaction in which the argon
nucleus is broken into an particle
reaction process can be written as
and an unknown entity, which we designate as
Conservation of the total number of nucleons gives
Conservation of the total electric charge yields
number of sulfur is
(b)
The reaction
, or
.
. The atomic
, or
, so the unknown entity must be sulfur
. The
.
can be written as
Since the total number of nucleons is conserved, we see that
Conservation of the total electric charge indicates that
unknown entity has
, so it must be a proton
, or
.
, or
. The
.
(c)
The reaction
can be written as
Conservation of the total number of nucleons reveals that the atomic mass number of the unknown
entity is given by
, or
. Conservation of the total electric
charge reveals that the atomic number is given by
, or
.
Potassium (K) has an atomic number
, so the unknown entity is the potassium nucleus
.
(d)
From the notation
, we see that the reaction is induced by a -ray photon, which has
. Therefore, the reaction process is
The unknown entity, then, has
, and
. The atomic number
corresponds to neon, so the unknown entity is neon
.
(b)
Answer:
REASONING During each reaction, both the total electric charge of the nucleons and the total
number of nucleons are conserved. For each type of nucleus that participates in a reaction, the atomic mass
number A specifies the number of nucleons, and the atomic number Z specifies the number of protons. We
will use these conserved quantities to calculate the atomic mass number and atomic number for each of the
unknown entities, thereby identifying them.
SOLUTION
(a)
In the reaction described by
?, a neutron
induces a reaction in which the argon
nucleus is broken into an particle
reaction process can be written as
and an unknown entity, which we designate as
Conservation of the total number of nucleons gives
Conservation of the total electric charge yields
number of sulfur is
, or
, or
, so the unknown entity must be sulfur
.
. The
.
. The atomic
(b)
The reaction
can be written as
Since the total number of nucleons is conserved, we see that
Conservation of the total electric charge indicates that
unknown entity has
, so it must be a proton
, or
.
, or
. The
.
(c)
The reaction
can be written as
Conservation of the total number of nucleons reveals that the atomic mass number of the unknown
entity is given by
, or
. Conservation of the total electric
charge reveals that the atomic number is given by
, or
.
Potassium (K) has an atomic number
, so the unknown entity is the potassium nucleus
.
(d)
From the notation
, we see that the reaction is induced by a -ray photon, which has
. Therefore, the reaction process is
The unknown entity, then, has
, and
. The atomic number
corresponds to neon, so the unknown entity is neon
.
(c)
Answer:
REASONING During each reaction, both the total electric charge of the nucleons and the total
number of nucleons are conserved. For each type of nucleus that participates in a reaction, the atomic mass
number A specifies the number of nucleons, and the atomic number Z specifies the number of protons. We
will use these conserved quantities to calculate the atomic mass number and atomic number for each of the
unknown entities, thereby identifying them.
SOLUTION
(a)
In the reaction described by
?, a neutron
induces a reaction in which the argon
nucleus is broken into an particle
reaction process can be written as
and an unknown entity, which we designate as
Conservation of the total number of nucleons gives
Conservation of the total electric charge yields
number of sulfur is
(b)
The reaction
, or
.
can be written as
Since the total number of nucleons is conserved, we see that
Conservation of the total electric charge indicates that
unknown entity has
.
. The atomic
, or
, so the unknown entity must be sulfur
. The
, so it must be a proton
, or
, or
.
.
. The
(c)
The reaction
can be written as
Conservation of the total number of nucleons reveals that the atomic mass number of the unknown
entity is given by
, or
. Conservation of the total electric
charge reveals that the atomic number is given by
, or
.
Potassium (K) has an atomic number
, so the unknown entity is the potassium nucleus
.
(d)
From the notation
, we see that the reaction is induced by a -ray photon, which has
. Therefore, the reaction process is
The unknown entity, then, has
, and
. The atomic number
corresponds to neon, so the unknown entity is neon
.
(d)
Answer:
REASONING During each reaction, both the total electric charge of the nucleons and the total
number of nucleons are conserved. For each type of nucleus that participates in a reaction, the atomic mass
number A specifies the number of nucleons, and the atomic number Z specifies the number of protons. We
will use these conserved quantities to calculate the atomic mass number and atomic number for each of the
unknown entities, thereby identifying them.
SOLUTION
(a)
In the reaction described by
?, a neutron
induces a reaction in which the argon
nucleus is broken into an particle
reaction process can be written as
and an unknown entity, which we designate as
Conservation of the total number of nucleons gives
Conservation of the total electric charge yields
number of sulfur is
(b)
The reaction
, or
.
can be written as
Since the total number of nucleons is conserved, we see that
Conservation of the total electric charge indicates that
unknown entity has
.
. The atomic
, or
, so the unknown entity must be sulfur
. The
, so it must be a proton
, or
.
, or
. The
.
(c)
The reaction
can be written as
Conservation of the total number of nucleons reveals that the atomic mass number of the unknown
entity is given by
, or
. Conservation of the total electric
charge reveals that the atomic number is given by
, or
.
Potassium (K) has an atomic number
, so the unknown entity is the potassium nucleus
.
(d)
From the notation
, we see that the reaction is induced by a -ray photon, which has
. Therefore, the reaction process is
The unknown entity, then, has
, and
. The atomic number
corresponds to neon, so the unknown entity is neon
.
REASONING During each reaction, both the total electric charge of the nucleons and the total number of
nucleons are conserved. For each type of nucleus that participates in a reaction, the atomic mass number A specifies
the number of nucleons, and the atomic number Z specifies the number of protons. We will use these conserved
quantities to calculate the atomic mass number and atomic number for each of the unknown entities, thereby
identifying them.
SOLUTION
(a)
In the reaction described by
?, a neutron
induces a reaction in which the argon
nucleus is
broken into an particle
written as
and an unknown entity, which we designate as
Conservation of the total number of nucleons gives
of the total electric charge yields
, or
, so the unknown entity must be sulfur
(b)
The reaction
can be written as
, or
. Conservation
. The atomic number of sulfur is
.
Since the total number of nucleons is conserved, we see that
total electric charge indicates that
, or
, so it must be a proton
. The reaction process can be
, or
. Conservation of the
. The unknown entity has
.
(c)
The reaction
can be written as
Conservation of the total number of nucleons reveals that the atomic mass number of the unknown entity is
given by
, or
. Conservation of the total electric charge reveals that the
atomic number is given by
, or
. Potassium (K) has an atomic number
, so the unknown entity is the potassium nucleus
.
(d)
From the notation
, we see that the reaction is induced by a -ray photon, which has
Therefore, the reaction process is
The unknown entity, then, has
, and
corresponds to neon, so the unknown entity is neon
*18.
. The atomic number
.
During a nuclear reaction, an unknown particle is absorbed by a copper
nucleus, and the reaction products are
, a neutron, and a proton. What are the name, atomic number, and nucleon number of the nucleus formed
temporarily when the copper
nucleus absorbs the unknown particle?
.
*19.
Consider the induced nuclear reaction
,
. The atomic masses are
and
,
. Determine the energy (in MeV) released by
the reaction.
Answer:
13.6 MeV
Section 32.3 Nuclear Fission, Section 32.4 Nuclear Reactors
20.
Determine the atomic number Z, the atomic mass number A, and the element X for the unknown species
following reaction for the fission of uranium
in the
:
Consult the periodic table on the inside of the back cover of the text as needed.
21.
When a
nucleus fissions, about 200 MeV of energy is released. What is the ratio of this
energy to the rest energy of the uranium nucleus?
Answer:
REASONING The rest energy of the uranium nucleus can be found by taking the atomic mass of the
atom, subtracting the mass of the 92 electrons, and then using the fact that 1 u is equivalent to 931.5 MeV (see Section
31.3). According to Table 31.1, the mass of an electron is
nucleus is found, the desired ratio can be calculated.
SOLUTION The mass of
. Once the rest energy of the uranium
is 235.043 924 u. Therefore, subtracting the mass of the 92 electrons, we have
The energy equivalent of this mass is
Therefore, the ratio is
22.
How many neutrons are produced when
fissions in the following way?
23.Neutrons released by a fission reaction must be slowed by collisions with the moderator nuclei before the neutrons can
cause further fissions. Suppose a 1.5-MeV neutron leaves each collision with 65% of its incident energy. How many
collisions are required to reduce the neutron's energy to at least 0.040 eV, which is the energy of a thermal neutron?
Answer:
41 collisions
24.
The energy released by each fission within the core of a nuclear reactor is
. The number of
fissions occurring each second is
25.
Uranium
. Determine the power (in watts) that the reactor generates.
fissions into two fragments plus three neutrons:
neutron is 1.008 665 u and the mass of
mass of the two fragments?
Answer:
232.7851 u
*26.
is 235.043 924 u. If 225.0 MeV of energy is released, what is the total
When 1.0 kg of coal is burned, approximately
each
fission is
. The mass of a
of energy is released. If the energy released during
, how many kilograms of coal must be burned to produce the same energy as 1.0
kg of
*27.
Suppose that the
nucleus fissions into two fragments whose mass ratio is 0.32:0.68. With the aid of Figure
32.8, estimate the energy (in MeV) released during this fission.
Answer:
160 MeV
REASONING AND SOLUTION The binding energy per nucleon for a nucleus with
is about 7.6
MeV per nucleon, according to Figure 32.9 The nucleus fragments into two pieces of mass ratio 0.32 : 0.68. These
fragments thus have nucleon numbers
Using Figure 32.9 we can estimate the binding energy per nucleon for
nucleon for
is about 8.8 MeV, representing an increase of
there are 76 nucleons present, the energy released for
is
Similarly, for
. We find that the binding energy per
per nucleon. Since
, we see that the binding energy increases to 8.0 MeV per nucleon, the difference being
per nucleon. Since there are 163 nucleons present, the energy released for
The energy released per fission is
*28.
and
is
.
The water that cools a reactor core enters the reactor at
it does not turn to steam.) The core is generating
and leaves at
. (The water is pressurized, so
of power. Assume that the specific heat capacity of
water is
over the temperature range stated above, and find the mass of water that passes through the
core each second.
**29.
A nuclear power plant is 25% efficient, meaning that only 25% of the power it generates goes into producing
electricity. The remaining 75% is wasted as heat. The plant generates
releases
Answer:
1200 kg
of energy, how many kilograms of
watts of electric power. If each fission
are fissioned per year?
REASONING We first determine the total power generated (used and wasted) by the plant. Energy is power
times the time, according to Equation 6.10a, and given the energy, we can determine how many kilograms of
fissioned to produce this energy.
SOLUTION Since the power plant produces energy at a rate of
total power produced by the power plant is
are
when operating at 25 % efficiency, the
The energy equivalent of one atomic mass unit is given in the text (see Section 31.3) as
Since each fission produces
year
is
of energy, the total mass of
required to generate
for a
**30.When a nuclear reactor is in a critical state, the neutrons released in each fission trigger an average of exactly one
additional fission. If the average number of additional fissions triggered rises above one, the reactor enters a
supercritical state in which the fission rate and the power output grow very rapidly. A reactor in a critical state
has a power output of 25 kW. The reactor then enters a supercritical state in which each fission triggers an
average of 1.01 additional fissions. The average time for the neutrons released by one generation of fissions to
trigger the next generation of fissions is
. How much time elapses before the power output from a
single generation of fissions grows to 3300 MW (roughly the normal output of a commercial reactor)?
Section
32.5
Nuclear
Fusion
31.
Two deuterium atoms (
reaction:
) react to produce tritium (
) and hydrogen (
) according to the following fusion
What is the energy (in MeV) released by this deuterium–deuterium reaction?
Answer:
4.03 MeV
32.
In one type of fusion reaction a proton fuses with a neutron to form a deuterium nucleus:
The masses are
,
energy (in MeV) is released by this reaction?
, and
. The -ray photon is massless. How much
33.
The fusion of two deuterium nuclei
and a neutron
can yield a helium nucleus
. What is the energy (in MeV) released in this
reaction?
Answer:
3.3 MeV
34.
Tritium (
) is a rare isotope of hydrogen that can be produced by the following fusion reaction:
(a)Determine the atomic mass number A, the atomic number Z, and the names X and Y of the unknown
particles.
(b)Using the masses given in the reaction, determine how much energy (in MeV) is released by this reaction.
*35.
One proposed fusion reaction combines lithium
with deuterium
to give helium
. How many kilograms of lithium would be needed to supply the energy needs
of one household for a year, estimated to be
Answer:
?
REASONING To find the energy released per reaction, we follow the usual procedure of determining how
much the mass has decreased because of the fusion process. Once the energy released per reaction is determined, we
can determine the mass of lithium
needed to produce
SOLUTION The reaction and the masses are shown below:
.
The mass defect is, therefore,
. Since 1 u is equivalent to 931.5 MeV,
the released energy is 21 MeV, or since the energy equivalent of one atomic mass unit is gven in Section 31.3 as
,
In 1.0 kg of lithium
, there are
Therefore, 1.0 kg of lithium
would produce an energy of
If the energy needs of one household for a year is estimated to be
*36.
, then the amount of lithium required is
In Example 6 it was determined that 17.6 MeV of energy is released when the following fusion reaction occurs:
Ignore relativistic effects and determine the kinetic energies of the neutron and the particle.
*37.
Deuterium
is an attractive fuel for fusion reactions because it is abundant in the oceans, where about 0.015% of
the hydrogen atoms in the water
are deuterium atoms.
(a)How many deuterium atoms are there in one kilogram of water?
Answer:
deuterium atoms
(b)If each deuterium nucleus produces about 7.2 MeV in a fusion reaction, how many kilograms of water
would be needed to supply the energy needs of the United States for one year, estimated to be
?
Answer:
**38.The proton–proton cycle thought to occur in the sun consists of the following sequence of reactions:
In these reactions
is a positron
, is a neutrino
, and is a gamma ray
photon
. Note that reaction (3) uses two
nuclei, which are formed by two reactions of type (1) and
two reactions of type (2). Verify that the proton–proton cycle generates about 25 MeV of energy. The atomic masses
are
(1.007 825 u),
(2.014 102 u),
(3.016 030 u), and
(4.002 603 u). Be sure to account for the fact that
there are two electrons in two hydrogen atoms, whereas there is one electron in a single deuterium
mass of one electron is 0.000 549 u.
atom. The
Section 32.6 Elementary Particles
39.The main decay mode for the negative pion is
. Find the energy (in MeV) released in this decay.
Consult Table 32.3 for rest energies and assume that the rest energy for
Answer:
33.9 MeV
MeV.
40.
A neutral pion
produced in a high-energy particle experiment moves at a
speed of 0.780 c. After a very short time, it decays into two -ray photons. One of the -ray photons has an energy of
192 MeV. What is the energy (in MeV) of the second -ray photon? Take relativistic effects into account.
41.The
particle has a charge of
and contains one quark and one antiquark.
Which
quarks
can
the
particle
not contain?
(a)
Answer:
The
particle does not contain u, c, or t quarks.
(b)Which antiquarks can the particle not contain?
Answer:
The
particle does not contain , , or antiquarks.
42.
In addition to its rest energy, a moving proton
has kinetic energy. This proton collides with a stationary
proton (p), and the reaction forms a stationary neutron (n), a stationary proton (p), and a stationary pion (
),
according to the following reaction:
. The rest energy of each proton is 938.3 MeV, and the
rest energy of the neutron is 939.6 MeV. The rest energy of the pion is 139.6 MeV. What is the kinetic energy (in
MeV) of the moving proton?
Suppose a neutrino is created and has an energy of 35 MeV.
43.
(a)Assuming the neutrino, like the photon, has no mass and travels at the speed of light, find the momentum of
the neutrino.
Answer:
REASONING The momentum of a photon is given in the text as
(see the discussion
leading to Equation 29.6). This expression applies to any massless particle that travels at the speed of light.
In particular, assuming that the neutrino has no mass and travels at the speed of light, it applies to the
neutrino. Once the momentum of the neutrino is determined, the de Broglie wavelength can be calculated
from Equation 29.6
.
SOLUTION
(a) The momentum of the neutrino is, therefore,
where we have used the fact that
(see Section 31.3).
(b)According to Equation 29.6, the de Broglie wavelength of the neutrino is
(b)Determine the de Broglie wavelength of the neutrino.
Answer:
REASONING The momentum of a photon is given in the text as
(see the discussion
leading to Equation 29.6). This expression applies to any massless particle that travels at the speed of light.
In particular, assuming that the neutrino has no mass and travels at the speed of light, it applies to the
neutrino. Once the momentum of the neutrino is determined, the de Broglie wavelength can be calculated
from Equation 29.6
.
SOLUTION
(a) The momentum of the neutrino is, therefore,
where we have used the fact that
(see Section 31.3).
(b)According to Equation 29.6, the de Broglie wavelength of the neutrino is
REASONING The momentum of a photon is given in the text as
(see the discussion leading to
Equation 29.6). This expression applies to any massless particle that travels at the speed of light. In particular,
assuming that the neutrino has no mass and travels at the speed of light, it applies to the neutrino. Once the
momentum of the neutrino is determined, the de Broglie wavelength can be calculated from Equation 29.6
.
SOLUTION
(a) The momentum of the neutrino is, therefore,
where we have used the fact that
(see Section 31.3).
(b)According to Equation 29.6, the de Broglie wavelength of the neutrino is
*44.
Review Conceptual Example 7 as background for this problem. An electron and its antiparticle annihilate each
other, producing two -ray photons. The kinetic energies of the particles are negligible. Determine the magnitude of
the momentum of each photon.
*45.Review Conceptual Example 5 as background for this problem. An energetic proton is fired at a stationary proton. For
the reaction to produce new particles, the two protons must approach each other to within a distance of about
. The moving proton must have a sufficient speed to overcome the repulsive Coulomb force. What
must be the minimum initial kinetic energy (in MeV) of the proton?
Answer:
0.18 MeV
Copyright © 2012 John Wiley & Sons, Inc. All rights reserved.