Names:________________________________________________________________________________
Physics 103 Team Session for Chapter 27, 29, & 30 - Solutions
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1) The dark fringe for m = 0 in a Young’s double-slit experiment is located at an angle of ϴ = 15 . What is the angle that
locates the dark fringe for m = 1?
REASONING In a Young’s double-slit experiment a dark fringe is located at an angle  that is determined
according to

sin   m 
1
2
 d
m  0, 1, 2, 3, ...
(27.2)
where  is the wavelength of the light and d is the separation of the slits. Here, neither  nor d is known.
However, we know the angle for the dark fringe for which m = 0. Using this angle and m = 0 in Equation 27.2
will allow us to determine the ratio /d, which can then be used to find the angle that locates the dark fringe for
m = 1.
SOLUTION Applying Equation 27.2 to the dark fringes for m = 0 and m = 1, we have
  d
sin 0  0 
1
2
and
  d
sin 1  1 
1
2
Dividing the expression for sin 1 by the expression for sin 0 gives
 
 
1
1
sin 1
2 d

 3.0
1 
sin 0
0
2 d
The angle that locates the dark fringe for m = 1 can now be found:
sin 1  3.0sin 0
1  sin 1 3.0sin 0   sin 1 3.0sin15  51
or
Alternate solution: Solve for the term λ/d using the angle you know and m=0. Then use that value to solve for
the unknown angle when m=1.
2) A flat screen is placed 4.5 m from a double-slit diffraction grating. The separation on the screen between the central
bright fringe and the first-order bright fringe is 0.037 m. The light passing through the slits has a wavelength of 490 nm.
What is the separation between the slits?
REASONING The slit separation d is given by Equation 27.1 with m  1; namely d   / sin  . As shown in
Example 1 in the text, and using the equation you used in lab, the angle  is given by   tan 1 ( y / L ) .
SOLUTION The angle  is
  tan 1
FG 0.037 m IJ  0.47
H 4.5 m K
Therefore, the slit separation d is

490  10 –9 m
d 


sin 
sin 0.47 
6.0  10 –5 m
-19
3) The work function of a metal surface is 4.8 x 10 J. The maximum speed of the electrons emitted from the surface is
5
va = 7.3 x 10 m/s when the wavelength of the light is λa. When the wavelength is λb, the max speed of the electrons is
5
vb = 5.0 x 10 m/s. What are the two wavelengths λa and λb?
REASONING According to Equation 29.3, the relation between the photon energy, the maximum kinetic
energy of an ejected electron, and the work function of a metal surface is
hf  KE max  W0
Photon
energy
Maximum
kinetic energy
of ejected
electron
Work
function
Equation 16.1 relates the frequency f of a photon to its wavelength  via f = c/, where c is the speed of light in
a vacuum. The maximum kinetic energy KEmax is related to the mass m and maximum speed vmax of the ejected
electron by KEmax = 21 mv 2 (Equation 6.2). With these substitutions, Equation 29.3 becomes
hc 1 2
hf  KEmax  W0
or
 mv
 W0
(1)
 2 max
SOLUTION Solving Equation (1) for the wavelength gives
hc
1 2
mvmax  W0
2
6.63  1034 J  s  3.00  108 m/s 

A 

2
1 9.11  1031 kg 7.30  105 m/s  4.80  10 19 J


2
2.75  107 m
6.63  1034 J  s  3.00  108 m/s 

B 

2
1 9.11  1031 kg 5.00  105 m/s  4.80  10 19 J


2
3.35  107 m
2
4) Light is shining perpendicularly on the surface of earth with an intensity of 680 W/m . Assuming that all the photons in
the beam of light has the same wavelength of 730 nm, determine the number of photons per second per square m that
reach the earth. (Hint: Power is equal to Energy/time, so Watts = Joules/second)
REASONING The intensity S = 680 W/m2 of the photons is equal to the total amount of energy delivered by
the photons per second per square meter of surface area. Therefore, the intensity of the photons is equal to the
energy E of one photon multiplied by the number N of photons that reach the surface of the earth per second per
square meter:
S  NE
(1)
The energy E of each photon is given by E  hf (Equation 29.2), where h = 6.63×10−34 J·s is Planck’s constant
c
and f is the frequency of the photon. We will use f  (Equation 16.1) to determine the frequency f of the

photons from their wavelength  and the speed c of light in a vacuum.
SOLUTION Solving Equation (1) for N, we obtain N = S/E. Substituting E  hf (Equation 29.2) into this
result yields
N
S
S

E hf
(2)
Substituting f 
c
(Equation 16.1) into Equation (2), we find that




680 W/m 2 730 109 m
S
S
S
N



hf
 c  hc
6.63 1034 J  s 3.00 108 m/s
h 




 2.5 1021 photons/  s  m 2 
5) Using Bohr’s model of energy levels, for a hydrogen atom,
(a) what is the ionization energy for an electron in the n=4 excited state?
(b) what is the ionization energy for an electron in the ground state?
REASONING The ionization energy for a given state is the energy needed to remove the electron completely
from the atom. The removed electron has no kinetic energy and no electric potential energy, so its total energy
is zero.
The ionization energy for a given excited state is less than the ionization energy for the ground state. In the
excited state the electron already has part of the energy necessary to achieve ionization, so less energy is
required to ionize the atom from the excited state than from the ground state.
SOLUTION
a.The energy of the nth state in the hydrogen atom is given by Equation 30.13 as En   13.6 eV  Z 2 / n2 .
When n =  , E  0 J , and when n = 4, E4   13.6 eV 1 / 42   0.850 eV . The difference in energies
between these
two states is the ionization energy:
Ionization energy = E – E4 = 0.850 eV = 1.36 x 10-19 J
2
b.
In the same manner, it can be shown that the ionization energy for the
n = 1 state is 13.6 eV or 2.18 x 10-18 J.
6) Which of the following subshell configurations are not allowed? For those not allowed, give the physics reason why.
a) 3s
1
2
b)2d
c)3s
4
8
d)4p
e)5f
12
REASONING
The orbital quantum number
can have any integer value from 0 up to
n – 1. If, for example, n = 4, can have the values 0, 1, 2, and 3.
For a given value of the orbital quantum number , the magnetic quantum number m can have any integer
value, including 0, from – to + . For instance, if = 2, m can have the values –2, –1, 0 ,+1, and +2.
SOLUTION Of the five subshell configurations, three are not allowed. The ones that are not allowed, and the
reasons they are not allowed, are:
(b)
The principal quantum number is n = 2 and the orbital quantum number is = 2 (the d
subshell). Since must be less than n, this subshell configuration is not permitted.
(c)
This subshell has 4 electrons. However, according to the Pauli exclusion principle, only two
electrons can be in the s subshell ( = 0). Therefore, this subshell configuration is not allowed.
(d)
This subshell has 8 electrons. However, according to the Pauli exclusion principle, only six
electrons can be in the p subshell ( = 1). Therefore, this subshell configuration is not allowed.