Chapter 34
Wave-Particle Duality and Quantum Physics
Conceptual Problems
1
•
[SSM] The quantized character of electromagnetic radiation is
observed by (a) the Young double-slit experiment, (b) diffraction of light by a
small aperture, (c) the photoelectric effect, (d) the J. J. Thomson cathode-ray
experiment.
Determine the Concept The Young double-slit experiment and the diffraction of
light by a small aperture demonstrated the wave nature of electromagnetic
radiation. J. J. Thomson’s experiment showed that the rays of a cathode-ray tube
were deflected by electric and magnetic fields and therefore must consist of
electrically charged particles. Only the photoelectric effect requires an
explanation based on the quantization of electromagnetic radiation. (c ) is
correct.
2
••
Two monochromatic light sources, A and B, emit the same number of
photons per second. The wavelength of A is λA = 400 nm, and the wavelength of
B is λB = 600 nm. The power radiated by source B (a) is equal to the power of
source A, (b) is less than the power of source A, (c) is greater than the power of
source A, (d) cannot be compared to the power from source A using the available
data.
B
Determine the Concept Since the power radiated by a source is the energy
radiated per unit area and per unit time, it is directly proportional to the energy.
The energy radiated varies inversely with the wavelength ( E = hc λ ); i.e., the
longer the wavelength, the less energy is associated with the electromagnetic
radiation. (b ) is correct.
3
•
[SSM] The work function of a surface is φ. The threshold
wavelength for emission of photoelectrons from the surface is equal to (a) hc/φ,
(b) φ/hf, (c) hf/φ, (d) none of above.
Determine the Concept The work function is equal to the minimum energy
required to remove an electron from the material. A photon that has that energy
also has the threshold wavelength required for photoemission. Thus, hf = φ . In
addition, c = fλ . It follows that hc λt = φ , so λt = hc φ and
1
(a )
is correct.
2
Chapter 34
4
••
When light of wavelength λ1 is incident on a certain photoelectric
cathode, no electrons are emitted, no matter how intense the incident light is. Yet,
when light of wavelength λ2 < λ1 is incident, electrons are emitted, even when the
incident light has low intensity. Explain this observation.
Determine the Concept We can use Einstein’s photoelectric equation to explain
this observation.
Einstein’s photoelectric equation is:
Because f = c/λ:
K max = hf − φ
where φ , called the work function, is a
characteristic of the particular metal.
K max =
We’re given that when light of
wavelength λ1 is incident on a
certain photoelectric cathode; no
electrons are emitted independently
of the intensity of the incident light.
Hence, we can conclude that:
hc
When light of wavelength λ2 < λ1 is
incident, electrons are emitted,
electrons are emitted independently
of the intensity of the incident light.
Hence, we can conclude that:
hc
λ1
λ2
hc
λ
−φ
−φ < 0⇒
hc
−φ > 0 ⇒
hc
λ1
λ2
<φ
>φ
Thus, photons with wavelengths greater than the threshold wavelength λ t = c f t ,
where ft is the threshold frequency, do not have enough energy to eject an electron
from the metal.
5
•
True or false:
(a) The wavelength of an electron’s matter wave varies inversely with the
momentum of the electron.
(b) Electrons can undergo diffraction.
(c) Neutrons can undergo diffraction.
(a) True. The de Broglie wavelength of an electron is given by λ = h p .
(b) True
(c) True
Wave-Particle Duality and Quantum Physics
3
6
•
If the wavelength of an electron is equal to the wavelength of a proton,
then (a) the speed of the proton is greater than the speed of the electron, (b) the
speeds of the proton and the electron are equal, (c) the speed of the proton is less
than the speed of the electron, (d) the energy of the proton is greater than the
energy of the electron, (e) Both (a) and (d) are correct.
Determine the Concept If the de Broglie wavelengths of an electron and a proton
are equal, their momenta must be equal. Because mp > me, vp < ve. (c ) is correct.
7
•
A proton and an electron have equal kinetic energies. It follows that
the wavelength of the proton is (a) greater than the wavelength of the electron,
(b) equal to the wavelength of the electron, (c) less than the wavelength of the
electron.
Determine the Concept The kinetic energy of a particle can be expressed, in
terms of its momentum, as K = p 2 2m . We can use the equality of the kinetic
energies and the fact that me < mp to determine the relative sizes of their de
Broglie wavelengths.
Express the equality of the kinetic
energies of the proton and electron in
terms of their momenta and masses:
pp2
2mp
=
pe2
2me
Use the de Broglie relation for the
wavelength of matter waves to
obtain:
h2
h2
⇒ mp λ2p = me λ2e
=
2mp λ2p 2me λ2e
Because me < mp:
λ2p < λ2e and λe > λp
and
(c )
is correct.
8
•
The parameter x is the position of a particle, <x> is the expectation
value of x, and P(x) is the probability density function. Can the expectation value
of x ever equal a value of x for which P(x) is zero? If yes, give a specific
example. If no, explain why not.
Determine the Concept Yes. Consider a particle in a one-dimensional box of
length L that is on the x axis on the interval 0 <x L. The wave function for a
particle in the n = 2 state (the lowest state above the ground state) is given
x⎞
2
⎛
byψ 2 ( x ) =
sin ⎜ 2π ⎟ (see Figure 31-13). The expectation value of x is L/2
L ⎝ L⎠
and P(L/2) = 0 (see Figure 31-14).
4
Chapter 34
9
••
It was once believed that if two identical experiments are done on
identical systems under the same conditions, the results must be identical. Explain
how this statement can be modified so that it is consistent with quantum physics.
Determine the Concept According to quantum theory, the average value of many
measurements of the same quantity will yield the expectation value of that
quantity. However, any single measurement may differ from the expectation
value.
10 ••
A six-sided die has the numeral 1 painted on three sides and the
numeral 2 painted on the other three sides. (a) What is the probability of a 1
coming up when the die is thrown? (b) What is the expectation value of the
numeral that comes up when the die is thrown? (c) What is the expectation value
of the cube of the numeral that comes up when the die is thrown?
Determine the Concept The probability of a particular event occurring is the
number of ways that event can occur divided by the number of possible outcomes.
The expectation value, on the other hand, is the average value of the experimental
results.
(a) Find the probability of a 1
coming up when the die is thrown:
P(1) =
3
1
=
6
2
3 ×1 + 3 × 2
= 1 .5
6
(b) Find the average value of a large
number of throws of the die:
n =
(c) The expectation value of the cube
of the numeral that comes up when
the die is thrown is:
3 × 13 + 3 × 2 3
n =
= 4.5
6
Estimation and Approximation
11
•• [SSM] During an advanced physics lab, students use X rays to
measure the Compton wavelength, λC. The students obtain the following
wavelength shifts
λ2 – λ1 as a function of scattering angle θ :
θ
λ2 − λ1
45º
75º
90º
135º
180º
0.647 pm 1.67 pm 2.45 pm 3.98 pm 4.95 pm
Use their data to estimate the value for the Compton wavelength. Compare this
number with the accepted value.
Wave-Particle Duality and Quantum Physics
Picture the Problem From the Compton-scattering equation we have
λ2 − λ1 = λC (1 − cosθ ) , where λC = h mec is the Compton wavelength. Note that
this equation is of the form y = mx + b provided we let y = λ2 − λ1 and
x = 1 − cosθ. Thus, we can linearize the Compton equation by plotting
Δλ = λ2 − λ1 as a function of 1− cosθ . The slope of the resulting graph will yield
an experimental value for the Compton wavelength.
(a) The spreadsheet solution is shown below. The formulas used to calculate the
quantities in the columns are as follows:
Cell
Formula/Content
Algebraic Form
A3
45
θ (deg)
B3 1 − cos(A3*PI()/180)
1 − cosθ
C3
6.47E^−13
Δλ = λ2 − λ1
θ
1− cosθ
λ2−λ1
(deg)
45
75
90
135
180
0.293
0.741
1.000
1.707
2.000
6.47E−13
1.67E−12
2.45E−12
3.98E−12
4.95E−12
The following graph was plotted from the data shown in the above table. Excel’s
″Add Trendline″ was used to fit a linear function to the data and to determine the
regression constants.
6.0E-12
Δλ = a(1 − cosθ) + b
5.0E-12
a = 2.48 × 10 −12 m
Δλ, m
4.0E-12
b = −1.03 × 10 −13 m
3.0E-12
2.0E-12
1.0E-12
0.0E+00
0.0
0.5
1.0
1 − cosθ
1.5
2.0
5
6
Chapter 34
From the trend line we note that the
experimental value for the Compton
wavelength λC,exp is:
λC,exp = 2.48 pm
The Compton wavelength is given
by:
λC =
h
hc
=
me c me c 2
Substitute numerical values and
evaluate λC:
λC =
1240 eV ⋅ nm
= 2.43 pm
5.11 × 10 5 eV
Express the percent difference
between λC and λC,exp:
% diff =
=
λC,exp − λexp λC,exp
=
−1
λexp
λexp
2.48 pm
−1 ≈ 2%
2.43 pm
12 ••
Students in a physics lab are trying to determine the value of Planck’s
constant h, using a photoelectric apparatus similar to the one shown in Figure 342. The students are using a helium–neon laser that has a tunable wavelength as the
light source. The data that the students obtain for the maximum electron kinetic
energies are
λ
Kmax
544 nm
594 nm
604 nm
612 nm
633 nm
0.360 eV 0.199 eV 0.156 eV 0.117 eV 0.062 eV
(a) Using a spreadsheet program or graphing calculator, plot Kmax versus light
frequency. (b) Use the graph to estimate the value of Planck’s constant. (Note:
You may wish to use a feature of your spreadsheet program or graphing
calculator to obtain the best straight-line fit to the data.) (c) Compare your result
with the accepted value for Planck’s constant.
Picture the Problem From Einstein’s photoelectric equation we have
K max = hf − φ , which is of the form y = mx + b , where the slope is h and the Kmaxintercept is the work function. Hence we should plot a graph of Kmax versus f in
order to obtain a straight line whose slope will be an experimental value for
Planck’s constant.
(a) The spreadsheet solution is shown below. The formulas used to calculate the
quantities in the columns are as follows:
Cell Formula/Content Algebraic Form
A3
544
λ (nm)
B3
0.36
Kmax (eV)
Wave-Particle Duality and Quantum Physics
C3
D3
E3
λ
(nm)
544
594
604
612
633
λ (m)
c/λ
A3*10^−19
3*10^8/C3
B3*1.6*10^−19
Kmax
(eV)
0.36
0.199
0.156
0.117
0.062
λ
(m)
5.44E−07
5.94E−07
6.04E−07
6.12E−07
6.33E−07
Kmax (J)
f = c/λ
(Hz)
5.51E+14
5.05E+14
4.97E+14
4.90E+14
4.74E+14
Kmax
(J)
5.76E−20
3.18E−20
2.50E−20
1.87E−20
9.92E−21
The following graph was plotted from the data shown in the above table. Excel’s
″Add Trendline″ was used to fit a linear function to the data and to determine the
regression constants.
7.E-20
6.E-20
5.E-20
K max = af + b
a = 6.19 × 10 −34 J ⋅ s
K max (J)
b = −2.83 × 10 −19 J
4.E-20
3.E-20
2.E-20
1.E-20
0.E+00
4.6E+14
4.8E+14
5.0E+14
5.2E+14
5.4E+14
5.6E+14
f (Hz)
(b) From the trend line we note that
the experimental value for Planck’s
constant is:
(c) Express the percent difference
between hexp and h:
hexp = 6.19 ×10−34 J ⋅ s
% diff =
h − hexp
= 1−
hexp
h
h
−34
6.19 × 10 J ⋅ s
= 1−
≈ 6 .6 %
6.63 × 10 −34 J ⋅ s
7
8
Chapter 34
13 ••
The cathode that was used by the students in the experiment described
in Problem 12 is constructed from one of the following metals:
Metals
Tungsten Silver Potassium Cesium
Work function 4.58 eV 2.4 eV
2.1 eV
1.9 eV
Determine which metal composes the cathode by using the same data given in
Problem 12. (a) Using a spreadsheet program or graphing calculator, plot Kmax
versus frequency. (b) Use the graph to estimate the value of the work function
based on the students’ data. (Note: You may wish to use a feature of your
spreadsheet program or graphing calculator to obtain the best straight-line fit to
the data.) (c) Which metal was most likely used for the cathode in their
experiment?
Picture the Problem From Einstein’s photoelectric equation we have
K max = hf − φ , which is of the form y = mx + b , where the slope is h and the
Kmax-intercept is the work function. Hence we should plot a graph of Kmax versus f
in order to obtain a straight line whose intercept will be an experimental value for
the work function.
(a) The spreadsheet solution is shown below. The formulas used to calculate the
quantities in the columns are as follows:
Cell Formula/Content Algebraic Form
A3
544
λ (nm)
B3
0.36
Kmax (eV)
C3
A3*10^−19
λ (m)
D3
3*10^8/C3
c/λ
E3 B3*1.6*10^−19
Kmax (J)
λ
Kmax
(nm) (eV)
544 0.36
594 0.199
604 0.156
612 0.117
633 0.062
λ
(m)
5.44E−07
5.94E−07
6.04E−07
6.12E−07
6.33E−07
f = c/λ
(Hz)
5.51E+14
5.05E+14
4.97E+14
4.90E+14
4.74E+14
Kmax
(J)
5.76E−20
3.18E−20
2.50E−20
1.87E−20
9.92E−21
Wave-Particle Duality and Quantum Physics
9
The following graph was plotted from the data shown in the above table. Excel’s
″Add Trendline″ was used to fit a linear function to the data and to determine the
regression constants.
7.E-20
6.E-20
5.E-20
K max = af + b
a = 6.19 × 10 −34 J ⋅ s
K max (J)
b = −2.83 × 10 −19 J
4.E-20
3.E-20
2.E-20
1.E-20
0.E+00
4.6E+14
4.8E+14
5.0E+14
5.2E+14
5.4E+14
5.6E+14
f (Hz)
(b) From the trend line we note that
the experimental value for the work
function φ is:
φexp = 2.83 × 10 −19 J ×
1eV
1.602 × 10 −19 J
= 1.77 eV
(c) The value of φexp = 1.77 eV is closest to the work function for cesium .
The Particle Nature of Light: Photons
14 •
Find the photon energy in electron volts for light of wavelength
(a) 450 nm, (b) 550 nm, and (c) 650 nm.
Picture the Problem We can use E = hc/λ to find the photon energy when we are
given the wavelength of the radiation.
(a) Express the photon energy as a
function of wavelength and evaluate
E for λ = 450 nm:
(b) For λ = 550 nm:
E=
E=
hc
λ
=
1240 eV ⋅ nm
= 2.76 eV
450 nm
1240 eV ⋅ nm
= 2.25 eV
550 nm
10
Chapter 34
(c) For λ = 650 nm:
E=
1240 eV ⋅ nm
= 1.91 eV
650 nm
15 •
Find the photon energy in electron volts for an electromagnetic wave
of frequency (a) 100 MHz in the FM radio band and (b) 900 kHz in the AM radio
band.
Picture the Problem We can find the photon energy for an electromagnetic wave
of a given frequency f from E = hf where h is Planck’s constant.
(a) For f = 100 MHz:
(
)(
)
)(
)
E = hf = 6.626 × 10 −34 J ⋅ s 100 × 10 6 s −1
1eV
= 6.626 × 10 − 26 J ×
1.602 × 10 −19 J
= 4.14 × 10 −7 eV
(b) For f = 900 kHz:
(
E = hf = 6.626 ×10 −34 J ⋅ s 900 × 10 3 s −1
1eV
= 5.963 ×10 −28 J ×
1.602 ×10 −19 J
= 3.72 × 10 −9 eV
16 •
What are the frequencies of photons that have the following energies
(a) 1.00 eV, (b) 1.00 keV, and (c) 1.00 MeV?
Picture the Problem The energy of a photon, in terms of its frequency, is given
by E=hf.
(a) Express the frequency of a
photon in terms of its energy and
evaluate f for E = 1.00 eV:
f =
(b) For E = 1.00 keV:
f =
1.00 eV
E
=
h 4.136 ×10 −15 eV ⋅ s
= 2.42 ×1014 Hz
1.00 keV
4.136 × 10 −15 eV ⋅ s
= 2.42 × 1017 Hz
(c) For E = 1.00 MeV:
f =
1.00 MeV
4.136 ×10 −15 eV ⋅ s
= 2.42 × 10 20 Hz
Wave-Particle Duality and Quantum Physics
11
17 •
Find the photon energy in electron volts if the wavelength is
(a) 0.100 nm (about 1 atomic diameter) and (b) 1.00 fm (1 fm = 10–15 m, about 1
nuclear diameter).
Picture the Problem We can use E = hc/λ to find the photon energy when we are
given the wavelength of the radiation.
hc
The energy of a photon as a function
of its wavelength is given by:
E=
(a) Substitute numerical values and
evaluate E for λ = 0.100 nm:
E=
1240 eV ⋅ nm
= 12.4 keV
0.100 nm
(b) Substitute numerical values and
evaluate E for λ =1.00 fm = 1.00 ×
10−6 nm:
E=
1240 eV ⋅ nm
= 1.24 GeV
1.00 × 10 −6 nm
λ
18 ••
The wavelength of red light emitted by a 3.00-mW helium–neon laser
is 633 nm. If the diameter of the laser beam is 1.00 mm, what is the density of
photons in the beam? Assume that the intensity is uniformly distributed across
the beam.
Picture the Problem We can express the density of photons in the beam as the
number of photons per unit volume. The number of photons per unit volume is, in
turn, the ratio of the power of the laser to the energy of the photons and the
volume occupied by the photons emitted in one second is the product of the crosssectional area of the beam and the speed at which the photons travel, i.e., the
speed of light.
Express the density of photons in the
beam as a function of the number of
photons emitted per second and the
volume occupied by those photons:
ρ=
N
V
Relate the number of photons
emitted per second to the power of
the laser and the energy of the
photons:
N=
P Pλ
=
E hc
Express the volume containing the
photons emitted in one second as a
function of the cross sectional area of
the beam:
V = Ac
12
Chapter 34
Substitute for N and V and simplify
to obtain:
ρ=
Pλ
Pλ
4 Pλ
= 2 π 2 =
2
hc A hc 4 d
πhc 2 d 2
(
)
where d is the diameter of the laser
beam.
Substitute numerical values and evaluate ρ:
ρ=
π (6.626 × 10
4(3.00 mW )(633 nm )
−34
)(
)
J ⋅ s 2.998 × 10 m/s (1.00 mm )
8
2
2
= 4.06 × 1013 m −3
19 •
[SSM] Lasers used in a telecommunications network typically
produce light that has a wavelength near 1.55 μm. How many photons per second
are being transmitted if such a laser has an output power of 2.50 mW?
Picture the Problem The number of photons per second is the ratio of the power
of the laser to the energy of the photons.
Relate the number of photons
emitted per second to the power of
the laser and the energy of the
photons:
N=
Substitute numerical values and
evaluate N:
N=
P P Pλ
=
=
E hc hc
λ
(2.50 mW )(1.55 μm )
(6.626 ×10
−34
)(
J ⋅ s 2.998 × 10 8 m/s
)
= 1.95 × 1016 s −1
The Photoelectric Effect
20 •
The work function for tungsten is 4.58 eV. (a) Find the threshold
frequency and wavelength for the photoelectric effect to occur when
monochromatic electromagnetic radiation is incident on the surface of a sample of
tungsten. Find the maximum kinetic energy of the electrons if the wavelength of
the incident light is (b) 200 nm and (c) 250 nm.
Picture the Problem The threshold wavelength and frequency for emission of
photoelectrons is related to the work function of a metal through φ = hf t = hc λt .
hc
We can use Einstein’s photoelectric equation K max =
− φ to find the maximum
λ
kinetic energy of the electrons for the given wavelengths of the incident light.
Wave-Particle Duality and Quantum Physics
(a) Express the threshold frequency
in terms of the work function for
tungsten and evaluate ft:
Using c = fλ, express the threshold
wavelength in terms of the threshold
frequency and evaluate λt:
(b) Using Einstein’s photoelectric
equation, relate the maximum kinetic
energy of the electrons to their
wavelengths and evaluate Kmax:
ft =
φ
h
=
13
4.58 eV
4.136 × 10 −15 eV ⋅ s
= 1.107 × 1015 Hz = 1.11 × 1015 Hz
λt =
c 2.998 × 108 m/s
=
= 271 nm
f t 1.107 × 1015 Hz
K max = E − φ = hf − φ =
=
hc
λ
−φ
1240 eV ⋅ nm
− 4.58 eV
200 nm
= 1.62 eV
(c) Evaluate Kmax for λ = 250 nm:
K max =
1240 eV ⋅ nm
− 4.58 eV
250 nm
= 0.380 eV
21 •
When monochromatic ultraviolet light that has a wavelength equal to
300 nm is incident on a sample of potassium, the emitted electrons have
maximum kinetic energy of 2.03 eV. (a) What is the energy of an incident
photon? (b) What is the work function for potassium? (c) What would be the
maximum kinetic energy of the electrons if the incident electromagnetic radiation
had a wavelength of 430 nm? (d) What is the maximum wavelength of incident
electromagnetic radiation that will result in the photoelectric emission of electrons
by a sample of potassium?
Picture the Problem (a) We can use the Einstein equation for photon energy to
find the energy of an incident photon. (b) and (c) We can use the photoelectric
equation to relate the work function for potassium to the maximum energy of the
photoelectrons. (d) The maximum wavelength of incident electromagnetic
radiation that will result in the photoelectric emission of electrons can be found
from λt = hc φ .
(a) Use the Einstein equation for
photon energy to relate the energy of
the incident photon to its
wavelength:
E=
hc
λ
=
1240 eV ⋅ nm
= 4.13 eV
300 nm
14
Chapter 34
(b) Using Einstein’s photoelectric
equation, relate the work function for
potassium to the maximum kinetic
energy of the photoelectrons:
K max = E − φ
Solve for and evaluate φ:
φ = E − K max = 4.13 eV − 2.03 eV
= 2.10 eV
(c) Proceeding as in (b) with
E = hc λ gives:
hc
K max =
=
λ
−φ
1240 eV ⋅ nm
− 2.10 eV
430 nm
= 0.78 eV
(d) Express the maximum
wavelength as a function of
potassium’s work function and
evaluate λt:
λt =
hc
φ
=
1240 eV ⋅ nm
= 590 nm
2.10 eV
22 •
The maximum wavelength of electromagnetic radiation that will result
in the photoelectric emission of electrons from a sample of silver is 262 nm.
(a) Find the work function for silver. (b) Find the maximum kinetic energy of the
electrons if the incident radiation has a wavelength of 175 nm.
Picture the Problem (a) We can find the work function for silver using
φ = hc λt and (b) the maximum kinetic energy of the electrons using Einstein’s
photoelectric equation.
(a) Express the work function for
silver as a function of the maximum
wavelength of the electromagnetic
radiation that will result in
photoelectric emission of electrons:
φ=
Substitute numerical values and
evaluate φ :
φ=
hc
λt
1240 eV ⋅ nm
= 4.73 eV
262 nm
Wave-Particle Duality and Quantum Physics
(b) Using Einstein’s photoelectric
equation, relate the work function for
silver to the maximum kinetic energy
of the photoelectrons:
K max = E − φ =
Substitute numerical values and
evaluate Kmax:
K max =
hc
λ
15
−φ
1240 eV ⋅ nm
− 4.73 eV
175 nm
= 2.36 eV
23 •
The work function for cesium is 1.90 eV. (a) Find the minimum
frequency and maximum wavelength of electromagnetic radiation that will result
in the photoelectric emission of electrons from a sample of cesium. Find the
maximum kinetic energy of the electrons if the wavelength of the incident
radiation is (b) 250 nm and (c) 350 nm.
Picture the Problem We can find the minimum frequency and maximum
wavelength for cesium using φ = hf t = hc λt and the maximum kinetic energy of
the electrons using Einstein’s photoelectric equation.
(a) Use the Einstein equation for
photon energy to express the
maximum wavelength for cesium:
λt =
Substitute numerical values and
evaluate λt :
λt =
Use c = fλ to find the maximum
frequency:
ft =
hc
φ
1240 eV ⋅ nm
= 653 nm
1.90 eV
c
λt
=
2.998 × 108 m/s
653 nm
= 4.59 × 1014 Hz
(b) Using Einstein’s photoelectric
equation, relate the maximum kinetic
energy of the photoelectrons to the
wavelength of the incident light:
Evaluating Kmax for λ = 250 nm
gives:
K max =
K max =
hc
λ
−φ
1240 eV ⋅ nm
− 1.90 eV
250 nm
= 3.06 eV
16
Chapter 34
(c) Proceed as in Part (b) with
λ = 350 nm to obtain:
K max =
1240 eV ⋅ nm
− 1.90 eV
350 nm
= 1.64 eV
24 ••
When a surface is illuminated with electromagnetic radiation of
wavelength 780 nm, the maximum kinetic energy of the emitted electrons is
0.37 eV. What is the maximum kinetic energy if the surface is illuminated using
radiation of wavelength 410 nm?
Picture the Problem We can use Einstein’s photoelectric equation to find the
work function of this surface and then apply it a second time to find the maximum
kinetic energy of the photoelectrons when the surface is illuminated with light of
wavelength 410 nm.
hc
Use Einstein’s photoelectric equation
to relate the maximum kinetic energy
of the emitted electrons to their total
energy and the work function of the
surface:
K max =
The work function of the surface is
given by:
φ = E − K max =
Substitute numerical values and
evaluate φ:
φ=
Substitute for φ and λ in equation (1)
and evaluate Kmax:
K max =
λ
−φ
(1)
hc
λ
− K max
1240 eV ⋅ nm
− 0.37 eV = 1.22 eV
780 nm
1240 eV ⋅ nm
− 1.22 eV
410 nm
= 1.80 eV
Compton Scattering
25 •
Find the shift in wavelength of photons scattered by free stationary
electrons at θ = 60°. (Assume that the electrons are initially moving with
negligible speed and are virtually free of (unattached to) any atoms or molecules.)
Picture the Problem We can calculate the shift in wavelength using the Compton
h
(1 − cosθ ) .
relationship Δλ =
me c
Wave-Particle Duality and Quantum Physics
The shift in wavelength is given by:
Δλ =
17
h
(1 − cos θ )
me c
Substitute numerical values and evaluate Δλ:
Δλ =
6.626 ×10 −34 J ⋅ s
(1 − cos60°) = 1.2 pm
9.109 × 10 −31 kg 2.998 ×108 m/s
(
)(
)
26
•
When photons are scattered by electrons in a carbon sample, the
shift in wavelength is 0.33 pm. Find the scattering angle. (Assume that the
electrons are initially moving with negligible speed and are virtually free of
(unattached to) any atoms or molecules.)
Picture the Problem We can calculate the scattering angle using the Compton
h
relationship Δλ =
(1 − cosθ ) .
me c
Using the Compton scattering
equation, relate the shift in
wavelength to the scattering angle:
Δλ =
Solving for θ yields:
h
(1 − cosθ )
me c
⎛
⎝
θ = cos −1 ⎜1 −
me c
⎞
Δλ ⎟
h
⎠
Substitute numerical values and evaluate θ :
⎛
θ = cos −1 ⎜⎜1 −
⎝
(9.109 ×10
)(
)
⎞
kg 2.998 ×108 m/s
(0.33 pm )⎟⎟ = 30°
−34
6.626 ×10 J ⋅ s
⎠
−31
27 •
The photons in a monochromatic beam are scattered by electrons. The
wavelength of the photons that are scattered at an angle of 135° with the direction
of the incident photon beam is 2.3 percent less than the wavelength of the incident
photons. What is the wavelength of the incident photons?
Picture the Problem We can calculate the shift in wavelength using the Compton
h
relationship Δλ =
(1 − cosθ ) .
me c
Express the wavelength of the
incident photons in terms of the
fractional change in wavelength:
Δλ
λ
= 0.023 ⇒ λ =
Δλ
0.023
18
Chapter 34
Using the Compton scattering
equation, relate the shift in
wavelength to the scattering angle:
Δλ =
Substituting for Δλ yields:
λ=
h
(1 − cosθ )
me c
h
(1 − cosθ )
0.023me c
Substitute numerical values and evaluate λ:
λ=
6.626 ×10 −34 J ⋅ s
(1 − cos135°) = 0.18 nm
(0.023) 9.109 ×10 −31 kg 2.998 ×108 m/s
(
)(
)
28 •
Compton used photons of wavelength 0.0711 nm. (a) What is the
energy of one of these photons? (b) What is the wavelength of the photons
scattered in the direction opposite to the direction of the incident photons?
(c) What is the energy of the photon scattered in this direction?
Picture the Problem We can use the Einstein equation for photon energy to find
the energy of both the incident and scattered photon and the Compton scattering
equation to find the wavelength of the scattered photon.
hc
1240 eV ⋅ nm
= 17.4 keV
0.0711 nm
(a) Use the Einstein equation for
photon energy to obtain:
E=
(b) Express the wavelength of the
scattered photon in terms of its prescattering wavelength and the shift in
its wavelength during scattering:
λ2 = λ1 + Δλ = λ1 +
λ1
=
h
(1 − cos θ )
me c
Substitute numerical values and evaluate λ2:
λ2 = 0.0711nm +
6.626 ×10 −34 J ⋅ s
(1 − cos180°) = 0.0760 nm
9.109 ×10 −31 kg 2.998 ×108 m/s
(
(c) Use the Einstein equation for
photon energy to obtain:
)(
)
E=
hc
λ2
=
1240 eV ⋅ nm
= 16.3 keV
0.0760 nm
29 •
For the photons used by Compton (see Problem 28), find the
momentum of the incident photon and the momentum of the photon scattered in
the direction opposite to the direction of the incident photons. Use the
Wave-Particle Duality and Quantum Physics
19
conservation of momentum to find the momentum of the recoil electron in this
case.
Picture the Problem Compton used X rays of wavelength 71.1 pm. Let the
direction the incident photon (and the recoiling electron) is moving be the positive
direction. We can use p = h/λ to find the momentum of the incident photon and
the conservation of momentum to find its momentum after colliding with the
electron.
Use the expression for the
momentum of a photon to find the
momentum of Compton’s photons:
Using the Compton scattering
equation, relate the shift in
wavelength to the scattering angle:
p1 =
h
λ1
=
6.626 × 10 −34 J ⋅ s
71.1pm
= 9.32 × 10 −24 kg ⋅ m/s
λ2 = λ1 + λC (1 − cos θ )
Substitute numerical values and evaluate λ2:
λ2 = 71.1 pm + (2.43 × 10 −12 m )(1 − cos180°) = 76.0 pm
Apply conservation of momentum
to obtain:
p1 = pe − p2 ⇒ pe = p1 − p2
Substitute for p1 and p2 and evaluate pe:
pe = 9.32 ×10
− 24
⎛ 6.626 ×10 −34 J ⋅ s ⎞
⎟⎟ = 1.80 ×10 −23 kg ⋅ m/s
kg ⋅ m/s − ⎜⎜ −
76.0 pm
⎝
⎠
30 ••
A beam of photons have a wavelength equal to 6.00 pm is scattered by
electrons initially at rest. A photon in the beam is scattered in a direction
perpendicular to the direction of the incident beam. (a) What is the change in
wavelength of the photon? (b) What is the kinetic energy of the electron?
Picture the Problem We can calculate the shift in wavelength using the Compton
h
(1 − cosθ ) = λC (1 − cosθ ) and use conservation of energy to
relationship Δλ =
me c
find the kinetic energy of the scattered electron.
20
Chapter 34
(a) Use the Compton scattering
equation to find the change in
wavelength of the photon:
Δλ = λC (1 − cos θ )
(
)
= 2.43 × 10 −12 m (1 − cos90°)
= 2.43 pm
hc
hc
(b) Use conservation of energy to
relate the change in the kinetic
energy of the electron to the
energies of the incident and
scattered photon:
ΔEe =
Find the wavelength of the scattered
photon:
λ2 = λ1 + Δλ = 6.00 pm + 2.43 pm
Substitute numerical values and
evaluate the kinetic energy of the
electron (equal to the change in its
energy since it was stationary prior
to the collision with the photon):
⎛ 1
1 ⎞
⎟⎟
ΔEe = 1240 eV ⋅ nm⎜⎜
−
⎝ 6.00 pm 8.43 pm ⎠
λ1
−
λ2
= 8.43 pm
= 60 keV
Electrons and Matter Waves
31 •
An electron is moving at 2.5 × 105 m/s. Find the electron’s de Broglie
wavelength.
Picture the Problem We can use its definition to find the de Broglie wavelength
of this electron.
Use its definition to express the de
Broglie wavelength of the electron in
terms of its momentum:
Substitute numerical values and
evaluate λ:
λ=
h
h
=
p me v
λ=
6.626 ×10 −34 J ⋅ s
(9.109 ×10 −31 kg )(2.5 ×105 m/s)
= 2.9 nm
32 •
An electron has a wavelength of 200 nm. Find (a) the magnitude of its
momentum and (b) its kinetic energy.
Wave-Particle Duality and Quantum Physics
21
Picture the Problem We can find the momentum of the electron from the de
1.226
nm , where K is in eV.
Broglie equation and its kinetic energy from λ =
K
(a) Use the de Broglie relation to
express the momentum of the
electron:
6.626 × 10 −34 J ⋅ s
p= =
200 nm
λ
(b) Use the electron wavelength
equation to relate the electron’s
wavelength to its kinetic energy:
λ=
Solving for K gives:
⎛ 1.226 eV1 2 nm ⎞
⎟⎟
K = ⎜⎜
200 nm
⎝
⎠
h
= 3.31× 10 −27 kg ⋅ m/s
1.226
nm
K
2
= 3.76 × 10 −5 eV
33 ••
An electron, a proton, and an alpha particle each have a kinetic energy
of 150 keV. Find (a) the magnitudes of their momenta and (b) their de Broglie
wavelengths.
Picture the Problem The momenta of these particles can be found from their
kinetic energies. The values for the electron, however, are only approximately
correct because at the given energy the speed of the electron is a significant
fraction of the speed of light. The solution presented is valid only in the nonrelativistic limit v << c. The de Broglie wavelengths of the particles are given by
λ = h/p.
(a) The momentum of a particle p, in
terms of its kinetic energy K, is given
by:
p = 2mK
Substitute numerical values and evaluate pe:
⎛
1.602 × 10 −19 C ⎞
⎟⎟
pe = 2me K = 2(9.109 × 10 −31 kg )⎜⎜150 keV ×
eV
⎝
⎠
= 2.092 × 10 −22 N ⋅ s = 2.09 × 10 − 22 N ⋅ s
22
Chapter 34
Substitute numerical values and evaluate pp:
⎛
1.602 ×10 −19 C ⎞
⎟⎟
pp = 2mp K = 2(1.673 ×10 −27 kg )⎜⎜150 keV ×
eV
⎝
⎠
= 8.967 ×10 −21 N ⋅ s = 8.97 ×10 − 21 N ⋅ s
Substitute numerical values and evaluate pα:
⎛
1.661×10 − 27 kg ⎞ ⎛
1.602 ×10 −19 C ⎞
⎟⎟ ⎜⎜150 keV ×
⎟⎟
pα = 2mα K = 2⎜⎜ 4 u ×
u
eV
⎝
⎠⎝
⎠
= 1.787 ×10 −20 N ⋅ s = 1.79 ×10 − 20 N ⋅ s
(b) The de Broglie wavelengths of
the particles are given by:
λ=
Substitute numerical values in
equation (1) and evaluate λe:
λe =
h
p
(1)
h
6.626 × 10 −34 J ⋅ s
=
pe 2.092 × 10 − 21 N ⋅ s
= 3.17 pm
Substitute numerical values in
equation (1) and evaluate λp:
λp =
h
6.626 ×10 −34 J ⋅ s
=
pp 8.967 ×10 −21 N ⋅ s
= 73.9 fm
Substitute numerical values in
equation (1) and evaluate λα:
h
6.63 × 10 −34 J ⋅ s
=
λα =
pα 1.79 × 10 −20 N ⋅ s
= 37.0 fm
34 •
A neutron in a reactor has kinetic energy of approximately 0.020 eV.
Calculate the wavelength of this neutron.
Picture the Problem The wavelength associated with a particle of mass m and
1240 eV ⋅ nm
.
kinetic energy K is given by λ =
2mc 2 K
Wave-Particle Duality and Quantum Physics
Substituting numerical values
yields:
35
•
23
1240 eV ⋅ nm
= 0.20 nm
2(940 MeV )(0.020 eV )
λ=
Find the wavelength of a proton that has a kinetic energy of 2.00 MeV.
Picture the Problem The wavelength associated with a particle of mass m and
1240 eV ⋅ nm
.
kinetic energy K is given by λ =
2mc 2 K
Substituting numerical values
yields:
λ=
1240 eV ⋅ nm
2(938 MeV )(2.00 MeV )
= 20.2 fm
36 •
What is the kinetic energy of a proton whose wavelength is
(a) 1.00 nm and (b) 1.00 fm?
Picture the Problem We can solve Equation 34-15 ( λ =
1240 eV ⋅ nm
) for the
2mc 2 K
kinetic energy of the proton and use the rest energy of a proton mc2 = 938 MeV to
simplify our computation.
(1240 eV ⋅ nm )2
Solving Equation 34-15 for the
kinetic energy of the proton gives:
K=
(a) Substitute numerical values and
evaluate K for λ = 1.00 nm:
2
(
1240 eV ⋅ nm )
K=
2
2(938 MeV )(1.00 nm )
2mc 2 λ2
= 0.820 meV
(b) Evaluate K for λ = 1.00 fm:
K=
(1240 eV ⋅ nm )2
2
2(938 MeV )(1.00 fm )
= 820 MeV
37 •
The kinetic energy of the electrons in the electron beam in a run of
Davisson and Germer’s experiment was 54 eV. Calculate the wavelength of the
electrons in the beam.
Picture the Problem If K is in electron volts, the wavelength of a particle is
1.226
nm provided K is in eV.
given by λ =
K
24
Chapter 34
Evaluate λ for K = 54 eV:
λ=
1.226
1.226
nm =
nm = 0.17 nm
K
54
38 •
The distance between Li+ and Cl– ions in a LiCl crystal is 0.257 nm.
Find the energy of electrons that have a wavelength equal to this spacing.
1.226
nm , where K is in eV, to find the
K
energy of electrons whose wavelength is λ.
Picture the Problem We can use λ =
1.226
nm
K
Relate the wavelength of the
electrons to their kinetic energy:
λ=
Solving for K yields:
⎛ 1.226 nm ⋅ eV1 2 ⎞
⎟⎟
K = ⎜⎜
λ
⎝
⎠
Substituting numerical values
and evaluating K gives:
⎛ 1.226 nm ⋅ eV1 2 ⎞
⎟⎟ = 22.8 eV
K = ⎜⎜
⎝ 0.257 nm ⎠
2
2
39 •
[SSM] An electron microscope uses electrons that have energies
equal to 70 keV. Find the wavelength of these electrons.
Picture the Problem We can approximate the wavelength of 70-keV electrons
1.226
nm , where K is in eV. This solution is, however, only
using λ =
K
approximately correct because at the given energy the speed of the electron is a
significant fraction of the speed of light. The solution presented is valid only in
the non-relativistic limit v << c.
Relate the wavelength of the
electrons to their kinetic energy:
λ=
Substitute numerical values and
evaluate λ:
λ=
1.23
nm
K
1.226
70 × 10 3 eV
nm = 4.6 pm
40 •
What is the de Broglie wavelength of a neutron that has a speed of
1.00 × 106 m/s?
Picture the Problem We can use its definition to calculate the de Broglie
wavelength of a neutron with speed 106 m/s.
Wave-Particle Duality and Quantum Physics
Use its definition to express the de
Broglie wavelength of the neutron:
λn =
h
h
=
pn mn vn
Substitute numerical values and
evaluate λn:
λn =
6.626 × 10 −34 J ⋅ s
(1.675 ×10 −27 kg )(1.00 ×106 m/s)
25
= 0.396 pm
A Particle in a Box
41 ••
(a) Find the energy of the ground state (n = 1) and the first two excited
states of a neutron in a one-dimensional box of length L = 1.00 ×10–15 m = 1.00
fm (about the diameter of an atomic nucleus). Make an energy-level diagram for
this system. Calculate the wavelength of electromagnetic radiation emitted when
the neutron makes a transition from (b) n = 2 to n = 1, (c) n = 3 to n = 2, and
(d) n = 3 to n = 1.
h2
and
8mL2
the energies of the excited states using En = n 2 E1 . The wavelength of the
Picture the Problem We can find the ground-state energy using E1 =
electromagnetic radiation emitted when the neutron transitions from one state to
hc
).
another is given by the Einstein equation for photon energy ( E =
λ
h2
E1 =
8mn L2
(a) Express the ground-state energy:
Substitute numerical values and evaluate E1:
(6.626 ×10 J ⋅ s )
=
8(1.6749 ×10 kg )(1.00 ×10
−34
E1
− 27
2
−15
1eV
= 204.53 MeV
−19
m 1.602 ×10 J
)
2
= 205 MeV
Find the energies of the first two
excited states:
E 2 = 2 2 E1 = 4(204.53 MeV )
= 818 MeV
and
E3 = 3 2 E1 = 9(204.53 MeV )
= 1.84 GeV
26
Chapter 34
The energy-level diagram for this
system is shown to the right:
E E1
n
25
5
16
4
9
4
1
0
3
2
1
hc 1240 eV ⋅ nm
=
ΔE
ΔE
(b) Relate the wavelength of the
electromagnetic radiation emitted
during a neutron transition to the
energy released in the transition:
λ=
For the n = 2 to n = 1 transition:
ΔE = E2 − E1 = 4 E1 − E1 = 3E1
Substitute numerical values and
evaluate λ2→1:
λ 2→1 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
3E1
3(204.53 MeV )
= 2.02 fm
(c) For the n = 3 to n = 2 transition:
ΔE = E3 − E2 = 9 E1 − 4 E1 = 5 E1
Substitute numerical values and
evaluate λ3→2:
λ3→2 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
5 E1
5(204.53 MeV )
= 1.21 fm
(d) For the n = 3 to n = 1 transition:
ΔE = E3 − E1 = 9 E1 − E1 = 8E1
Substitute numerical values and
evaluate λ3→1:
λ3→1 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
8 E1
8(204.53 MeV )
= 0.758 fm
(a) Find the energy of the ground state (n = 1) and the first two excited
42 ••
states of a neutron in a one-dimensional box of length 0.200 nm (about the
diameter of a H2 molecule). Calculate the wavelength of electromagnetic radiation
emitted when the neutron makes a transition from (b) n = 2 to n = 1, (c) n = 3 to
n = 2, and (d) n = 3 to n = 1.
Wave-Particle Duality and Quantum Physics
27
h2
and
Picture the Problem We can find the ground-state energy using E1 =
8mL2
the energies of the excited states using En = n 2 E1 . The wavelength of the
electromagnetic radiation emitted when the neutron transitions from one state to
hc
).
another is given by the Einstein equation for photon energy ( E =
λ
(a) Express the ground-state energy:
E1 =
h2
8mn L2
Substitute numerical values and evaluate E1:
E1 =
(6.626 × 10
8(1.6749 × 10
− 27
−34
J ⋅ s)
2
1 eV
= 5.1133 meV = 5.11 meV
−19
kg )(0.200 nm ) 1.602 × 10 J
Find the energies of the first two
excited states:
2
E 2 = 2 2 E1 = 4(5.1133 meV )
= 20.5 meV
and
E3 = 3 2 E1 = 9(5.1133 meV)
= 46.0 meV
hc 1240 eV ⋅ nm
=
ΔE
ΔE
(b) Relate the wavelength of the
electromagnetic radiation emitted
during a neutron transition to the
energy released in the transition:
λ=
For the n = 2 to n = 1 transition:
ΔE = E2 − E1 = 4 E1 − E1 = 3E1
Substitute numerical values and
evaluate λ2→1:
λ 2→1 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
3E1
3(5.1133 meV )
= 80.8 μm
(c) For the n = 3 to n = 2 transition:
ΔE = E3 − E2 = 9 E1 − 4 E1 = 5 E1
Substitute numerical values and
evaluate λ3→2:
λ3→ 2 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
5 E1
5(5.1133 meV )
= 48.5 μm
28
Chapter 34
(d) For the n = 3 to n = 1 transition:
ΔE = E3 − E1 = 9 E1 − E1 = 8E1
Substitute numerical values and
evaluate λ3→1:
λ3→1 =
1240 eV ⋅ nm
1240 eV ⋅ nm
=
8 E1
8(5.1133 meV )
= 30.3 μm
Calculating Probabilities and Expectation Values
43 ••
A particle is in the ground state of a one-dimensional box that has
length L. (The box has one end at the origin and the other end on the positive x
axis.) Determine the probability of finding the particle in the interval of length
Δx = 0.002L and centered at (a) 14 x = L, (b) x = 12 L, and (c) x = 43 L. (Because Δx
is very small you need not do any integration.)
Picture the Problem The probability of finding the particle in some range Δx is
ψ 2 dx . The interval Δx = 0.002L is so small that we can neglect the variation in
ψ(x) and just compute ψ2Δx.
Express the probability of finding the
particle in the interval Δx:
P = P(x )Δx = ψ 2 ( x )Δx
Express the wave function for a
particle in the ground state:
ψ 1 (x ) =
Substitute forψ 1 (x ) to obtain:
P=
(a) Evaluate P at x = 4L:
πx
2
sin
L
L
2 2 πx
sin
Δx
L
L
2⎛
πx ⎞
= ⎜ sin 2 ⎟(0.002 L )
L⎝
L⎠
⎛ πx ⎞
= 0.004 sin 2 ⎜ ⎟
⎝L⎠
P(4 L ) = 0.004 sin 2
π (4 L )
2L
= 0.004 sin (2π ) = 0
2
(b) Evaluate P at x = L/2:
P(L 2 ) = 0.004 sin 2
πL
2L
⎛π ⎞
= 0.004 sin 2 ⎜ ⎟ = 1
⎝2⎠
Wave-Particle Duality and Quantum Physics
(c) Evaluate P at x = 3L/4:
29
⎛ 3πL ⎞
P (3L 4 ) = 0.004 sin 2 ⎜
⎟
⎝ 4L ⎠
⎛ 3π ⎞
= 0.004 sin 2 ⎜ ⎟ = 0.002
⎝ 4 ⎠
44 ••
A particle is in the second excited state (n = 3) of a one-dimensional
box that has length L. (The box has one end at the origin and the other end on the
positive x axis.) Determine the probability of finding the particle in the interval of
length Δx = 0.002L and centered at (a) x = 13 L, (b) x = 12 L, and (c) x = 23 L.
(Because Δx is very small you need not do any integration.)
Picture the Problem The probability of finding the particle in some range dx is
ψ 2 dx . The interval Δx = 0.002L is so small that we can neglect the variation in
ψ(x) and just compute ψ 2Δx.
Express the probability of finding
the particle in the interval Δx:
P = P(x )Δx = ψ 2 ( x )Δx
Express the wave function for a
particle in its second excited
state:
ψ 2 (x ) =
Substitute for ψ 2 ( x ) to obtain:
2
2πx
sin
L
L
2 2 2πx
Δx
sin
L
L
2⎛
2πx ⎞
= ⎜ sin 2
⎟(0.002 L )
L⎝
L ⎠
⎛ 2πx ⎞
= 0.004 sin 2 ⎜
⎟
⎝ L ⎠
P=
(a) Evaluate P at x = L/3:
⎛ 2πL ⎞
P (L 3) = 0.004 sin 2 ⎜
⎟
⎝ 3L ⎠
⎛ 2π ⎞
= 0.004 sin 2 ⎜
⎟ = 0.003
⎝ 3 ⎠
(b) Evaluate P at x = L/2:
⎛ 2πL ⎞
P (L 2 ) = 0.004 sin 2 ⎜
⎟
⎝ 2L ⎠
= 0.004 sin 2 (π ) = 0
30
Chapter 34
(c) Evaluate P at x = 2L/3:
⎛ 4πL ⎞
P (2 L 3) = 0.004 sin 2 ⎜
⎟
⎝ 3L ⎠
⎛ 4π ⎞
= 0.004 sin 2 ⎜
⎟ = 0.003
⎝ 3 ⎠
45 ••
A particle is in the first excited (n = 2) state of a one-dimensional box
that has length L. (The box has one end at the origin and the other end on the
positive x axis.) Find (a) x and (b) x 2 .
Picture the Problem We’ll use f ( x ) = ∫ f ( x )ψ 2 ( x ) dx withψ n (x ) =
(a) Express ψ(x) for the n = 2 state:
Express x using the n = 2 wave
function:
Change variables by letting
2πx
. Then:
θ=
L
Substitute to obtain:
2 x 2 2πx
sin
dx
L
L
0
L
x =∫
L
L
2π
θ , dθ =
dx, dx =
dθ
2π
2π
L
and the limits on θ are 0 and 2π.
x=
L
2 ⎛ L ⎞
⎛ L
⎞
x = ∫ ⎜ θ ⎟ sin 2 θ ⎜
dθ ⎟
L 0 ⎝ 2π ⎠
⎝ 2π
⎠
wave function:
L
2π 2
2π
∫ θ sin
2
θ dθ
0
2π
L ⎡θ 2 θ sin 2θ cos 2θ ⎤
x =
−
−
2π 2 ⎢⎣ 4
4
8 ⎥⎦ 0
=
(b) Express x 2 using the n = 2
2
2πx
sin
L
L
ψ 2 (x ) =
=
Use a table of integrals to evaluate
the integral:
nπx
2
sin
.
L
L
L ⎡ 2 1 1⎤
L
π − + ⎥=
2 ⎢
2π ⎣
8 8⎦
2
2x2
2πx
sin 2
dx
L
L
0
L
x2 = ∫
Wave-Particle Duality and Quantum Physics
Change variables as in (a) and
substitute to obtain:
L
x
2
2
= ∫
L0
31
2
⎛ L ⎞
⎞
2 ⎛ L
dθ ⎟
⎜ θ ⎟ sin θ ⎜
⎝ 2π ⎠
⎝ 2π
⎠
2π
L2
=
θ 2 sin 2 θ dθ
3 ∫
4π 0
Use a table of integrals to evaluate the integral:
2π
x
2
L2 ⎡θ 3 ⎛ θ 2 1 ⎞
L2 ⎡ 4π 3 π ⎤
θ cos 2θ ⎤
⎜
⎟
=
− ⎥
sin
2
−
θ
=
−
−
⎥
⎢
4π 3 ⎣ 6 ⎜⎝ 4 8 ⎟⎠
4 ⎦0
4π 3 ⎢⎣ 3
2⎦
1 ⎞
⎛1
= L2 ⎜ − 2 ⎟ = 0.321L2
⎝ 3 8π ⎠
46 ••
A particle in a one-dimensional box that has length L is in the first
excited state (n = 2). (The box has one end at the origin and the other end on the
positive x axis.) (a) Sketch ψ 2(x) versus x for this state. (b) What is the
expectation value <x> for this state? (c) What is the probability of finding the
particle in some small region dx centered at x = L/2? (d) Are your answers for Part
(b) and Part (c) contradictory? If not, explain why your answers are not
contradictory.
Picture the Problem We’ll use f ( x ) = ∫ f ( x )ψ 2 ( x ) dx with
nπx
2
sin
. In Part (c) we’ll use P( x ) = ψ 22 ( x ) to determine the
L
L
probability of finding the particle in some small region dx centered at x = ½L.
ψ n (x ) =
ψ 2 (x ) =
Square both sides of the equation to
obtain:
ψ 22 ( x ) =
The graph of ψ 2(x) as a function of x
is shown to the right:
2
2πx
sin
L
L
2 2 2πx
sin
L
L
probability density
(a) Express the wave function for a
particle in its first excited state:
0
L
x
32
Chapter 34
(b) Express x using the n = 2 wave
function:
Change variables by letting
2πx
. Then:
θ=
L
Substitute in the expression for x
to obtain:
2 x 2 2πx
sin
dx
L
L
0
L
x =∫
L
θ,
2π
2π
dθ =
dx, and
L
L
dx =
dθ
2π
and the limits on θ are 0 and 2π.
x=
L
x =
=
Using a table of integrals, evaluate
the integral:
2 ⎛ L ⎞ 2 ⎛ L
⎞
dθ ⎟
⎜ θ ⎟ sin θ ⎜
∫
2
π
2
π
L0⎝
⎠
⎝
⎠
2π
L
2π
2
∫ θ sin
2
θ dθ
0
2π
L ⎡θ 2 θ sin 2θ cos 2θ ⎤
x =
−
−
8 ⎥⎦ 0
2π 2 ⎢⎣ 4
4
=
L ⎡ 2 1 1⎤
L
π − + ⎥=
2 ⎢
2π ⎣
8 8⎦
2
2
2πx
sin 2
L
L
(c) Express P(x):
P ( x ) = ψ 22 ( x ) =
Evaluate P(L/2):
2π L
⎛L⎞ 2
P⎜ ⎟ = sin 2
⋅
L 2
⎝2⎠ L
2
= sin 2 π = 0
L
Because P(L/2) = 0:
⎛L⎞
P⎜ ⎟dx = 0
⎝2⎠
(d) The answers to Parts (b) and (c) are not contradictory. Part (b) states that the
average value of measurements of the position of the particle will yield L/2 even
though the probability that any one measurement of position will yield this value
is zero.
Wave-Particle Duality and Quantum Physics
33
47 ••
A particle of mass m has a wave function given byψ ( x ) = Ae
,
where A and a are positive constants. (a) Find the normalization constant A.
(b) Calculate the probability of finding the particle in the region –a ≤ x ≤ a.
−x a
Picture the Problem We can find the constant A by applying the normalization
∞
condition ∫ψ 2 (x ) dx = 1 and finding the value for A that satisfies this condition. As
−∞
soon as we have found the normalization constant, we can calculate the
probability of the finding the particle in the region −a ≤ x ≤ a using
a
P = ∫ψ 2 ( x ) dx .
−a
(a) The normalization condition is:
∞
∫ψ (x ) dx = 1
2
−∞
Substitute ψ (x ) = Ae
−x a
:
∫ (Ae
∞
) dx = 2 A ∫ e
−∞
From integral tables:
∞
∫e
−αx
dx =
∞
2A
2
∫e
(b) Express the normalized wave
function:
The probability of finding the
particle in the region −a ≤ x ≤ a is:
−2 x a
dx
1
α
−2 x a
0
Solving for A yields:
2
0
0
Therefore:
∞
−x a 2
⎛a⎞
dx = 2 A2 ⎜ ⎟ = aA2 = 1
⎝2⎠
1
a
A=
ψ (x ) =
1 −x
e
a
a
P = ∫ψ
−a
a
a
2
(x ) dx = 2∫ 1 e −2 x a dx
0
a
a
=
2 −2 x a
e
dx = 1 − e −2 = 0.865
∫
a0
48 ••
A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L. A
particle in this box is in its ground state. Calculate the probability that the particle
will be found in the region (a) 0 < x < 12 L , (b) 0 < x < 13 L , and (c) 0 < x < 34 L .
34
Chapter 34
Picture the Problem The probability density for the particle in its ground state is
π
2
given by P (x ) = sin 2 x . We’ll evaluate the integral of P(x) between the limits
L
L
specified in (a), (b), and (c).
π
2
P(x ) = ∫ sin 2 xdx
L0
L
d
Express P(x) for 0 < x < d:
Change variables by letting θ =
π
L
x.
Then:
x=
L
π
dx =
Substitute to obtain:
θ , dθ =
L
π
π
L
dx, and
dθ
θ'
2
⎛L
⎞
P ( x ) = ∫ sin 2 θ ⎜ dθ ⎟
L0
⎝π
⎠
=
2
θ'
sin
π∫
2
θdθ
0
Use a table of integrals to evaluate
θ'
∫ sin
2
θ dθ :
θ'
2 ⎡θ sin 2θ ⎤
P( x ) = ⎢ −
4 ⎥⎦ 0
π ⎣2
(1)
0
(a) Noting that the limits on θ are 0
and π/2, evaluate equation (1) over
the interval 0 < x < ½L:
(b) Noting that the limits on θ are 0
and π/3, evaluate equation (1) over
the interval 0 < x < L/3:
P( x ) =
2 ⎡θ sin 2θ ⎤
−
4 ⎥⎦
π ⎢⎣ 2
π 2
=
0
= 0.500
π 3
2 ⎡θ sin 2θ ⎤
P(x ) = ⎢ −
π ⎣2
4 ⎥⎦ 0
2 ⎡ π sin 2π 3 ⎤
= ⎢ −
π ⎣6
4 ⎥⎦
=
1
3
−
= 0.196
3 4π
2 ⎡π ⎤
π ⎢⎣ 4 ⎥⎦
Wave-Particle Duality and Quantum Physics
(c) Noting that the limits on θ are 0
and 3π/4, Evaluate equation (1) over
the interval 0 < x < 3L/4:
2 ⎡θ sin 2θ ⎤
P( x ) = ⎢ −
π ⎣2
4 ⎥⎦
35
3π 4
0
2 ⎡ 3π sin 6π 4 ⎤
−
⎥⎦
4
π ⎢⎣ 8
3 1
= +
= 0.909
4 2π
=
49 ••
A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L. A
particle in this box is in its first excited state. Calculate the probability that the
particle will be found in the region (a) 0 < x < 12 L, (b) 0 < x < 13 L, and
(c) 0 < x < 43 L.
Picture the Problem The probability density for the particle in its first excited
2
2π
state is given by P ( x ) = sin 2
x . We’ll evaluate the integral of P(x) between
L
L
the limits specified in (a), (b), and (c).
2
2π
sin 2
xdx
∫
L0
L
d
Express P(x) for 0 < x < d:
P( x ) =
Change variables by letting
2π
θ=
x . Then:
L
L
2π
dx, and
θ , dθ =
2π
L
L
dx =
dθ
2π
Substitute to obtain:
P(x ) =
x=
=
Use a table of integrals to evaluate
θ'
2
∫ sin θ dθ :
θ'
2
⎛ L
⎞
sin 2 θ ⎜
dθ ⎟
∫
L0
⎝ 2π
⎠
1
π
θ'
∫ sin
2
θdθ
0
1 ⎡θ
P( x ) = ⎢
π ⎣2
θ'
−
sin 2θ ⎤
4 ⎥⎦ 0
(1)
0
(a) Noting that the limits on θ are 0
and π, evaluate equation (1) over the
interval 0 < x < ½L:
π
1 ⎡θ sin 2θ ⎤
1 ⎡π ⎤
= ⎢ ⎥
P( x ) = ⎢ −
⎥
4 ⎦0 π ⎣ 2 ⎦
π ⎣2
= 0.500
36
Chapter 34
(b) Noting that the limits on θ are 0
and 2π/3, evaluate equation (1) over
the interval 0 < x < L/3:
2π 3
1 ⎡θ sin 2θ ⎤
P( x ) = ⎢ −
π ⎣2
4 ⎥⎦ 0
1 ⎡ 2π sin 4π 3 ⎤
= ⎢
−
π⎣ 6
4 ⎥⎦
=
(c) Noting that the limits on θ are 0
and 3π/2, evaluate equation (1) over
the interval 0 < x < 3L/4:
1
3 2
+
= 0.402
3
4π
3π 2
1 ⎡θ sin 2θ ⎤
P(x ) = ⎢ −
4 ⎥⎦ 0
π ⎣2
=
1 ⎡ 3π ⎤
π ⎢⎣ 4 ⎥⎦
= 0.750
50 ••
The classical probability distribution function for a particle in a onedimensional box on the x axis in the region 0 < x < L is given by P(x) = 1/L. Use
this expression to show that x = 12 L and x 2 = 13 L2 for a classical particle in
the box.
Picture the Problem Classically, x = ∫ xP( x ) dx and x 2 = ∫ x 2 P ( x ) dx .
Evaluate x with P(x) = 1/L:
Evaluate x 2 with P(x) = 1/L:
L
⎡ x2 ⎤
x
x = ∫ dx = ⎢ ⎥ =
L
⎣ 2L ⎦ 0
0
L
x
L
L
⎡ x3 ⎤
x2
= ∫ dx = ⎢ ⎥ =
L
⎣ 3L ⎦ 0
0
L
2
1
2
1
3
L2
A one-dimensional box is on the x-axis in the region of 0 ≤ x ≤ L.
51 ••
(a) The wave functions for a particle in the box are given by
2
nπ x
ψ n (x ) =
sin
n = 1, 2, 3, . . .
L
L
For a particle in the nth state, show that x = 12 L and x 2 = L2/3 – L2/(2n2π2).
(b) Compare these expressions for x and x 2 , for n >> 1, with the expressions
for x and x 2 for the classical distribution of Problem 50.
Picture the Problem We’ll use f ( x ) = ∫ f ( x )ψ 2 ( x ) dx with
ψ n (x ) =
nπx
2
sin
to show that
L
L
L2
L2
L
2
x
=
=
and
.
x =
3 2n 2π 2
2
Wave-Particle Duality and Quantum Physics
2 x 2 nπx
sin
dx
L
L
L
(a) Express x for a particle in the
x =∫
nth state:
0
Change variables by letting θ =
nπx
.
L
Then:
L
nπ
L
θ , dθ =
dx, dx =
dθ
nπ
L
nπ
and the limits on θ are 0 and nπ.
x=
Substitute to obtain:
x =
2
L
nπ
⎛ L
⎞
∫ ⎜⎝ nπ θ ⎟⎠ sin
2
⎛ L
⎞
dθ ⎟
π
n
⎝
⎠
θ⎜
0
nπ
2L
= 2 2 ∫ θ sin 2 θ dθ
nπ 0
Use a table of integrals to evaluate the integral:
nπ
2 L ⎡θ 2 θ sin 2θ cos 2θ ⎤
x = 2 2⎢ −
−
4
8 ⎥⎦ 0
nπ ⎣4
2 L ⎡ n 2π 2 nπ sin 2nπ cos 2nπ 1 ⎤
2 L ⎡ n 2π 2 1 1 ⎤
−
−
+
=
− + ⎥
4
8
8 ⎥⎦ n 2π 2 ⎢⎣ 4
8 8⎦
n 2π 2 ⎢⎣ 4
L
=
2
=
2x2
nπx
sin 2
dx
L
L
0
L
Express x 2 for a particle in the
x2 = ∫
nth state:
Change variables by letting θ =
Then:
Substitute to obtain:
nπx
.
L
L
nπ
L
θ , dθ =
dx, dx =
dθ
L
nπ
nπ
and the limits on θ are 0 and nπ.
x=
x
2
2
=
L
nπ
∫
0
2
⎛ L ⎞
⎛ L
⎞
θ ⎟ sin 2 θ ⎜
dθ ⎟
⎜
⎝ nπ ⎠
⎝ nπ
⎠
nπ
=
2 L2
θ 2 sin 2 θ dθ
3 3 ∫
nπ 0
37
38
Chapter 34
Use a table of integrals to evaluate the integral:
nπ
x
2 L2 ⎡θ 3 ⎛ θ 2 1 ⎞
θ cos 2θ ⎤
2 L2 ⎡ n 3π 3 nπ ⎤
= 3 3 ⎢ − ⎜⎜ − ⎟⎟ sin 2θ −
=
−
⎥
n π ⎣ 6 ⎝ 4 8⎠
4 ⎦0
n 3π 3 ⎢⎣ 6
4 ⎥⎦
2
=
L2
L2
− 2 2
3 2n π
(b) For large values of n the result agrees with the classical value of L2/3 given in
Problem 50.
52
(a) Use a spreadsheet program or graphing calculator to plot x 2 as
••
a function of the quantum number n for the particle in the box described in
Problem 48 and for values of n from 1 to 100. Assume L = 1.00 m for your graph.
Refer to Problem 51. (b) Comment on the significance of any asymptotic limits
that your graph shows.
Picture the Problem (a) From Problem 51 we have x =
L
and
2
L2
L2
− 2 2 . A spreadsheet program was used to plot the following graph
3 2n π
2
of <x > as a function of n.
x2 =
0.34
0.33
0.32
0.30
2
<x >
0.31
0.29
0.28
0.27
0.26
0.25
1
10
19
28
37
46
55
n
64
73
82
91
100
Wave-Particle Duality and Quantum Physics
(b) As n → ∞, x
2
39
L2
→
3
53 ••
The wave functions for a particle of mass m in a one-dimensional box
of length L centered at the origin (so that the ends are at x = ± 12 L) are given by
ψ (x ) =
2
nπ x
cos
L
L
n = 1, 3, 5, 7, . . .
and
2
nπ x
sin
n = 2, 4, 6, 8, . . .
L
L
Calculate x and x 2 for the ground state (n = 1).
ψ (x ) =
Picture the Problem For the ground state, n = 1 and so we’ll evaluate
πx
2
cos .
f ( x ) = ∫ f ( x )ψ 2 ( x ) dx using ψ 1 (x ) =
L
L
Because ψ 12 ( x ) is an even function of
x, xψ 12 ( x ) is an odd function of x. It
follows that the integral of xψ 12 ( x )
between −L/2 and L/2 is zero. Thus:
L2
Express x 2 :
Change variables by letting θ =
x = 0 for all values of n.
x
πx
Then:
Substitute for x and dx to obtain:
L
.
2
π
2
=
x 2 cos 2 xdx
∫
L −L 2
L
L
θ , dθ =
π 2
x2 =
=
Use a trigonometric identity to
rewrite the integrand:
π
L
dθ
π
L
π
and the limits on θ are −π/2 and π/2.
x=
x
2
=
=
dx, dx =
2
2
⎛L ⎞
⎞
2 ⎛ L
⎜ θ ⎟ cos θ ⎜ dθ ⎟
∫
L −π 2 ⎝ π ⎠
⎝π
⎠
2 L2
π
3
2 L2
π3
π 2
∫π θ
−
2
cos 2 θ dθ
2
π 2
∫ θ (1 − sin θ )dθ
π
2
−
2
2
π 2
π 2
⎤
2 L2 ⎡
2
2
2
−
θ
d
θ
θ
sin
θ
d
θ
⎥
3 ⎢ ∫
∫
π ⎣⎢ −π 2
−π 2
⎦⎥
40
Chapter 34
Evaluate the second integral by looking it up in the tables:
π 2
x
2
θ cos 2θ ⎫⎤
2 L2 ⎡θ 3 ⎧θ 3 ⎛ θ 2 1 ⎞
= 3 ⎢ − ⎨ − ⎜⎜ − ⎟⎟ sin 2θ −
⎬⎥
π ⎢⎣ 3 ⎩ 6 ⎝ 4 8 ⎠
4 ⎭⎥⎦
−π
2
π 2
2 L2 ⎡θ 3 ⎛ θ 2 1 ⎞
θ cos 2θ ⎤
= 3 ⎢ + ⎜⎜ − ⎟⎟ sin 2θ +
⎥
π ⎣ 6 ⎝ 4 8⎠
4 ⎦ −π
=
2
2L ⎡π ⎛ π
1⎞
π cos π
+ ⎜⎜
− ⎟⎟ sin π +
3 ⎢
π ⎣ 48 ⎝ 16 8 ⎠
8
2
+
3
2
1⎞
π cos π ⎤
+ ⎜⎜
− ⎟⎟ sin π +
⎥
48 ⎝ 16 8 ⎠
8 ⎦
π3 ⎛π 2
2 L2 ⎡ π 3 π ⎤
1 ⎤
⎡1
= 3 ⎢ − ⎥ = L2 ⎢ − 2 ⎥
π ⎣ 24 4 ⎦
⎣12 2π ⎦
Remarks: The result differs from that of Example 34-8. Since we have shifted
the origin by Δx = L/2, we could have arrived at the above result, without
performing the integration, by subtracting (Δx)2 = L2/4 from x 2 .
54 ••
Calculate <x> and <x2> for the first excited state (n = 2) of the box
described in Problem 53.
Picture the Problem For the first excited state, n = 2, and so we’ll evaluate
2
2πx
sin
.
f ( x ) = ∫ f ( x )ψ 2 ( x ) dx using ψ 2 (x ) =
L
L
Because ψ 22 (x ) is an even function of
x, xψ 22 ( x ) is an odd function of x. It
follows that the integral of xψ 22 ( x )
between −L/2 and L/2 is zero. Thus:
L2
Express x 2 :
Change variables by letting θ =
Then:
x = 0
x2 =
2πx
.
L
2
2π
x 2 sin 2
xdx
∫
L −L 2
L
L
2π
L
θ , dθ =
dx, dx =
dθ
2π
L
2π
and the limits on θ are −π and π.
x=
Wave-Particle Duality and Quantum Physics
π
Substitute to obtain:
x
2
41
2
2 ⎛ L ⎞
⎛ L
⎞
= ∫ ⎜ θ ⎟ sin 2 θ ⎜
dθ ⎟
L −π ⎝ 2π ⎠
⎝ 2π
⎠
π
L2
=
θ 2 sin 2 θ dθ
3 ∫
4π −π
Evaluate the integral by looking it up in the tables:
π
x
2
θ cos 2θ ⎤
L2 ⎡θ 3 ⎛ θ 2 1 ⎞
=
− ⎜⎜ − ⎟⎟ sin 2θ −
⎥
3 ⎢
4π ⎣ 6 ⎝ 4 8 ⎠
4 ⎦ −π
=
π cos 2π
L2 ⎡ π 3 ⎛ π 2 1 ⎞
− ⎜⎜
− ⎟⎟ sin 2π −
3 ⎢
4π ⎣ 6 ⎝ 4 8 ⎠
4
+
1⎞
π cos 2π ⎤
− ⎜⎜
− ⎟⎟ sin 2π −
⎥
6 ⎝ 4 8⎠
4
⎦
π3 ⎛π2
=
L2 ⎡ π 3 π π 3 π ⎤
− +
−
4π 3 ⎢⎣ 6 4 6 4 ⎥⎦
=
L2 ⎡ π 3 π ⎤
1 ⎤
⎡1
− ⎥ = L2 ⎢ − 2 ⎥
3 ⎢
4π ⎣ 3 2 ⎦
⎣12 8π ⎦
Remarks: The result differs from that of Example 34-8. Since we have shifted
the origin by Δx = L/2, we could have arrived at the above result, without
performing the integration, by subtracting (Δx)2 = L2/4 from x 2 .
General Problems
55 •
[SSM] Photons in a uniform 4.00-cm-diameter light beam have
wavelengths equal to 400 nm and the beam has an intensity of 100 W/m2.
(a) What is the energy of each photon in the beam? (b) How much energy strikes
an area of 1.00 cm2 perpendicular to the beam in 1 s? (c) How many photons
strike this area in 1.00 s?
Picture the Problem We can use the Einstein equation for photon energy to find
the energy of each photon in the beam. The intensity of the energy incident on the
surface is the ratio of the power delivered by the beam to its delivery time. Hence,
we can express the energy incident on the surface in terms of the intensity of the
beam.
42
Chapter 34
(a) Use the Einstein equation for
photon energy to express the energy
of each photon in the beam:
Ephoton = hf =
Substitute numerical values and
evaluate Ephoton:
E photon =
hc
λ
1240 eV ⋅ nm
= 3.100 eV
400 nm
= 3.10 eV
(b) Relate the energy incident on a
surface of area A to the intensity of
the beam:
Substitute numerical values and
evaluate E:
E = IAΔt
E = (100 W/m 2 )(1.00 × 10 −4 m 2 )(1.00 s )
1eV
= 0.0100 J ×
1.602 × 10 −19 J
= 6.242 × 1016 eV = 6.24 × 1016 eV
(c) Express the number of photons
striking this area in 1.00 s as the ratio
of the total energy incident on the
surface to the energy delivered by
each photon:
N=
E
Ephoton
=
6.242 × 1016 eV
3.100 eV
= 2.08 × 1016
56 •
A 1-μg particle is moving with a speed of approximately 1 mm/s in a
one-dimensional box that has a length equal to 1 cm. Calculate the approximate
value of the quantum number n of the state occupied by the particle.
h2
. We can find
8mL2
n by solving this equation for n and substituting the particle’s kinetic energy for
En.
Picture the Problem The particle’s nth-state energy is En = n 2
Express the energy of the particle
when it is in its nth state:
En = n 2
h2
L
⇒n =
8mEn
2
8mL
h
Express the energy (kinetic) of the
particle:
En = 12 mv 2
Substitute for En and simplify to
obtain:
n=
2mvL
h
Wave-Particle Duality and Quantum Physics
Substitute numerical values and
evaluate n:
43
2(10 −9 kg )(1 × 10 −3 m/s )(10 −2 m )
n=
6.626 × 10 −34 J ⋅ s
≈ 3 × 1019
57 •
(a) For the particle and box of Problem 56, find Δx and Δpx, assuming
that these uncertainties are given by Δx/L = 0.01 percent and Δpx/px = 0.01
percent. (b) What is (ΔxΔp x ) h ?
Picture the Problem We can use the fact that the uncertainties are given by
Δx/L = 0.01 percent and Δpx/px = 0.01 percent to find Δx and Δp.
(
)
(a) Assuming that Δx/L = 0.01
percent, find Δx:
Δx = 10 −4 (L ) = 10 −4 10 −2 m = 1 μm
Assuming that Δpx/px = 0.01 percent,
find Δp:
Δp x = 10 −4 mv = 10 −4 10 −9 kg 10 −3 m/s
(b) Evaluate (ΔxΔp x ) h :
ΔxΔp x (1 μm )(10 −16 kg ⋅ m/s )
=
h
6.626 ×10 −34 J ⋅ s
(
)(
= 10 −16 kg ⋅ m/s
= 2 ×1011
58 •
In 1987, a laser at Los Alamos National Laboratory produced a flash
that lasted 1 × 10–12 s and had a power of 5 × 1015 W. Estimate the number of
emitted photons, assuming they all had wavelengths equal to 400 nm.
Picture the Problem We can estimate the number of emitted photons from the
ratio of the total energy in the flash to the energy of a single photon.
E
Letting N be the number of emitted
photons, express the ratio of the total
energy in the flash to the energy of a
single photon:
N=
Relate the energy in the flash to the
power produced:
E = PΔt
Express the energy of a single
photon as a function of its
wavelength:
Ephoton =
Ephoton
hc
λ
)
44
Chapter 34
Substituting for E and Ephoton and
simplifying gives:
N=
PΔ tλ
hc
Substitute numerical values and evaluate N:
N=
(5 × 10 W )(1× 10 s)(400 × 10 m ) ≈
(6.626 × 10 J ⋅ s )(2.998 × 10 m/s)
−12
15
−9
−34
8
1 × 10 22
59 •
You cannot ″see″ anything smaller than the wavelength of the wave
used to make the observation. What is the minimum energy of an electron needed
in an electron microscope to ″see″ an atom that has a diameter of about
0.1 nm?
Picture the Problem We can use the electron wavelength equation λ =
1.23
nm,
K
where K is in eV to find the minimum energy required to see an atom.
(
Relate the energy of the electron to
the size of an atom (the wavelength
of the electron):
1.23
1.23 eV1 2 ⋅ nm
λ=
nm ⇒ K =
λ2
K
provided K is in eV.
Substitute numerical values and
evaluate K:
(1.23 eV
K=
)
2
⋅ nm )
≈ 0.2 keV
(0.1nm )2
12
2
60 •
A common flea that has a mass of 0.008 g can jump vertically as high
as 20 cm. Estimate the de Broglie wavelength for the flea immediately after
takeoff.
Picture the Problem The flea’s de Broglie wavelength is λ = h p , where p is the
flea’s momentum immediately after takeoff. We can use a constant acceleration
equation to find the flea’s speed and, hence, momentum immediately after
takeoff.
h
h
=
p mv0
Express the de Broglie wavelength
of the flea immediately after takeoff:
λ=
Using a constant acceleration
equation, express the height the flea
can jump as a function of its takeoff
speed:
v 2 = v02 + 2aΔy
or, since v = 0 and a = −g,
v0 = 2 gΔy
Wave-Particle Duality and Quantum Physics
Substitute to obtain:
λ=
Substitute numerical values and
evaluate λ:
λ=
45
h
m 2 gΔy
(8 ×10
6.626 × 10 −34 J ⋅ s
−6
kg ) 2(9.81 m/s 2 )(0.2 m )
≈ 4 × 10 −29 m
61 ••
[SSM] A 100-W source radiates light of wavelength 600 nm
uniformly in all directions. An eye that has been adapted to the dark has a 7-mmdiameter pupil and can detect the light if at least 20 photons per second enter the
pupil. How far from the source can the light be detected under these rather
extreme conditions?
Picture the Problem We can relate the fraction of the photons entering the eye to
ratio of the area of the pupil to the area of a sphere of radius R. We can find the
number of photons emitted by the source from the rate at which it emits and the
energy of each photon which we can find using the Einstein equation.
Letting r be the radius of the pupil,
Nentering eye the number of photons per
second entering the eye, and Nemitted
the number of photons emitted by the
source per second, express the
fraction of the light energy entering
the eye at a distance R from the
source:
Solving for R yields:
N entering eye
N emitted
=
=
Aeye
4πR 2
πr 2
4πR 2
r2
=
4R 2
R=
r
2
N emitted
N entering eye
Find the number of photons emitted
by the source per second:
N emitted =
Using the Einstein equation, express
the energy of the photons:
Ephoton =
Substitute numerical values and
evaluate Ephoton:
Ephoton =
(1)
P
Ephoton
hc
λ
1240 eV ⋅ nm
= 2.07 eV
600 nm
46
Chapter 34
Substitute and evaluate Nemitted:
N emitted =
100 W
(2.07 eV )⎛⎜1.602 × 10 −19
⎝
J ⎞
⎟
eV ⎠
= 3.02 × 10 s −1
20
Substitute for Nemitted in equation (1)
and evaluate R:
R=
3.5 mm 3.02 × 10 20 s −1
2
20 s −1
≈ 7 × 10 3 km
62 ••
The diameter of the pupil of an eye under room-light conditions is
approximately 5 mm. Find the intensity of light that has a wavelength equal to
600 nm so that 1 photon per second passes through the pupil.
Picture the Problem The intensity of the light such that one photon per second
passes through the pupil is the ratio of the energy of one photon to the product of
the area of the pupil and time interval during which the photon passes through the
pupil. We’ll use the Einstein equation to express the energy of the photon.
Use its definition to relate the
intensity of the light to the energy of
a 600-nm photon:
I1 photon =
Using the Einstein equation, express
the energy of a 600-nm photon:
E1 photon =
Substitute for E1 photon to obtain:
I1 photon =
P E1 photon
=
A
AΔt
hc
λ
hc
λAΔt
Substitute numerical values and evaluate I1 photon:
I1 photon =
(1240 eV ⋅ nm )(1.602 × 10 −19 J/eV ) ≈
(600 nm ) ⎡⎢ π (5 × 10 −3 m )2 ⎤⎥(1s )
⎣4
2 × 10 −14 W/m 2
⎦
63 ••
A 100-W incandescent light bulb radiates 2.6 W of visible light
uniformly in all directions. (a) Find the intensity of the light from the bulb at a
distance of 1.5 m. (b) If the average wavelength of the visible light is 650 nm, and
counting only those photons in the visible spectrum, find the number of photons
per second that strike a surface that has an area equal to 1.0 cm2, is oriented so
that the line to the bulb is perpendicular to the surface, and is a distance of 1.5 m
from the bulb.
Wave-Particle Duality and Quantum Physics
47
Picture the Problem We can find the intensity at a distance of 1.5 m directly
from its definition. The number of photons striking the surface each second can be
found from the ratio of the energy incident on the surface to the energy of a
650-nm photon.
(a) Use its definition to express the
intensity of the light as a function of
distance from the light bulb:
I=
Substitute numerical data to obtain:
I=
P
P
=
A 4πR 2
2.6 W
4π (1.5 m )
2
= 91.96 mW/m 2
= 92 mW/m 2
IA
(b) Express the number of photons
per second that strike the surface as
the ratio of the energy incident on
the surface to the energy of a 650-nm
photon:
where A is the area of the surface.
Use the Einstein equation to express
the energy of the 650-nm photons:
E=
Substituting for E yields:
N=
N=
Ephoton
hc
λ
IAλ
hc
Substitute numerical values and evaluate N:
N=
(91.96 mW/m )(1.0 × 10
)
m 2 (650 nm )
= 3.0 × 10 4
(1240 eV ⋅ nm )⎛⎜1.602 × 10 −19 J ⎞⎟
eV ⎠
⎝
2
−4
64 ••
When light of wavelength λ1 is incident on the cathode of a
photoelectric tube, the maximum kinetic energy of the emitted electrons is 1.8 eV.
If the wavelength is reduced to 12 λ1, the maximum kinetic energy of the emitted
electrons is 5.5 eV. Find the work function φ of the cathode material.
Picture the Problem The maximum kinetic energy of the photoelectrons is
related to the frequency of the incident photons and the work function of the
cathode material through the Einstein equation. We can apply this equation to the
two sets of data and solve the resulting equations simultaneously for the work
function.
48
Chapter 34
Using the Einstein equation, relate
the maximum kinetic energy of the
emitted electrons to the frequency of
the incident photons and the work
function of the cathode material:
K max = hf − φ
Substitute numerical values for the
light of wavelength λ1:
1.8 eV =
Substitute numerical values for the
light of wavelength 12 λ1 :
5.5 eV =
Solve equations (1) and (2)
simultaneously for φ to obtain:
φ = 1.9 eV
=
hc
−φ
λ
hc
λ1
−φ
hc
1
2
λ1
−φ =
(1)
2hc
λ1
−φ
(2)
65 ••
An incident photon of energy Ei undergoes Compton scattering at an
angle of θ. Show that the energy Es of the scattered photon is given by
Es =
Ei
1 + Ei me c 2 (1 − cos θ )
(
)
Picture the Problem We can use the Einstein equation to express the energy of
the scattered photon in terms of its wavelength and the Compton scattering
equation to relate this wavelength to the scattering angle and the pre-scattering
wavelength.
Express the energy of the scattered
photon Es as a function of their
wavelength λs:
Es =
Express the wavelength of the
scattered photon as a function of
the scattering angle θ :
λs =
hc
λs
h
(1 − cos θ ) + λ
me c
where λ is the wavelength of the
incident photon.
Wave-Particle Duality and Quantum Physics
Substitute for λ and simplify to
obtain:
Es =
49
hc
h
(1 − cos θ ) + λ
me c
hc
λ
=
1+
=
hc
(1 − cos θ )
me c 2 λ
Ei
⎛ E ⎞
1 + ⎜⎜ i 2 ⎟⎟ (1 − cos θ )
⎝ me c ⎠
66 ••
A particle is confined to a one-dimensional box. While the particle
makes a transition from the state n to the state n – 1, radiation of 114.8 nm is
emitted. While the particle makes the transition from the state n – 1 to the state
n – 2, radiation of wavelength 147 nm is emitted. The ground-state energy of the
particle is 1.2 eV. Determine n.
Picture the Problem While we can work with either of the transitions described
in the problem statement, we’ll use the first transition in which radiation of
wavelength 114.8 nm is emitted. We can express the energy released in the
transition in terms of the difference between the energies in the two states and
solve the resulting equation for n.
Express the energy of the emitted
radiation as the particle goes from
the nth to n – 1 state:
ΔE = En − En −1
Express the energy of the particle in
nth state:
En = n 2 E1
Express the energy of the particle in
the n – 1 state:
En −1 = (n − 1) E1
Substitute for En and En−1 and
simplify to obtain:
ΔE = n 2 E1 − (n − 1) E1
Solving for n yields:
2
2
= (2n − 1)E1 =
n=
hc
1
+
2λE1 2
hc
λ
50
Chapter 34
Substitute numerical values and
evaluate n:
n=
1240 eV ⋅ nm
1
+ = 5
2(114.8 nm )(1.2 eV ) 2
67 ••
[SSM] The Pauli exclusion principle states that no more than one
electron may occupy a particular quantum state at a time. Electrons intrinsically
occupy two spin states. Therefore, if we wish to model an atom as a collection of
electrons trapped in a one-dimensional box, no more than two electrons in the box
can have the same value of the quantum number n. Calculate the energy that the
most energetic electron(s) would have for the uranium atom that has an atomic
number 92. Assume the box has a length of 0.050 nm and the electrons are in the
lowest possible energy states. How does this energy compare to the rest energy of
the electron?
Picture the Problem We can use the expression for the energy of a particle in a
well to find the energy of the most energetic electron in the uranium atom.
Relate the energy of an electron in
the uranium atom to its quantum
number n:
⎛ h2 ⎞
⎟
E n = n 2 ⎜⎜
2 ⎟
⎝ 8me L ⎠
Substitute numerical values and evaluate E92:
2
⎡
⎤
(
6.626 × 10 −34 J ⋅ s )
1eV
×
= 1.273 MeV
E92 = (92) ⎢
2
−19 ⎥
−31
⎢⎣ 8(9.109 × 10 kg )(0.050 nm ) 1.602 × 10 J ⎥⎦
2
= 1.3 MeV
The rest energy of an electron is:
⎞
2⎛
1eV
⎟ = 0.512 MeV
me c 2 = 9.109 × 10 −31 kg 2.998 × 108 m/s ⎜⎜
−19 ⎟
⎝ 1.602 × 10 J ⎠
(
)(
Express the ratio of E92 to mec2:
)
E92
1.273 MeV
=
≈ 2 .5
2
0.512 MeV
me c
The energy of the most energetic electron is approximately 2.5 times the restenergy of an electron.
68 ••
A beam of electrons that each have the same kinetic energy
illuminates a pair of slits separated by a distance 54 nm. The beam forms bright
and dark fringes on a screen located a distance 1.5 m beyond the two slits. The
arrangement is otherwise identical to that used in the optical two-slit interference
Wave-Particle Duality and Quantum Physics
51
experiment described in Chapter 33 and in Figure 33-7 and the fringes have the
appearance shown in Figure 34-18d. The bright fringes are found to be separated
by a distance of 0.68 mm. What is the kinetic energy of the electrons in the beam?
Picture the Problem We can express the kinetic energy of an electron in the
beam in terms of its momentum. We can use the de Broglie relationship to relate
the electron’s momentum to its wavelength and use the condition for constructive
interference to find λ.
n =1
Δy
d θ
θ
n=0
L
Δ r = d sin θ
Express the kinetic energy of an
electron in terms of its momentum:
Using the de Broglie relationship,
relate the momentum of an electron
to its wavelength:
Substitute for p in equation (1) to
obtain:
The condition for constructive
interference is:
K=
p=
K=
p2
2m
(1)
h
λ
h2
2mλ2
(2)
d sin θ
n
where d is the slit separation and
n = 0, 1, 2, …
d sin θ = nλ ⇒ λ =
Δy
L
For θ << 1, sinθ is also given by:
sin θ ≈
Substitute for sinθ to obtain:
λ=
Substitute for λ in equation (2) to
obtain:
n 2 L2 h 2
K=
2
2md 2 (Δy )
dΔy
nL
52
Chapter 34
Substitute numerical values (n = 1) and evaluate K:
2
2
2
(
1) (1.5 m ) (6.626 ×10 −34 J ⋅ s )
1eV
K=
=
2
2
−19
−31
2(9.109 ×10 kg )(54 nm ) (0.68 mm ) 1.602 ×10 J
2.5 keV
69 ••
When a surface is illuminated by light of wavelength λ, the maximum
kinetic energy of the emitted electrons is 1.20 eV. If the wavelength
λ′ = 0.800λ is used, the maximum kinetic energy increases to 1.76 eV. For
wavelength λ′ = 0.600λ, the maximum kinetic energy of the emitted electrons is
2.676 eV. Determine the work function of the surface and the wavelength λ.
Picture the Problem The maximum kinetic energy of the photoelectrons is
related to the frequency of the incident photons and the work function of the
illuminated surface through the Einstein equation. We can apply this equation to
either set of data and solve the resulting equations simultaneously for the work
function of the surface and the wavelength of the incident photons.
Using the Einstein equation, relate
the maximum kinetic energy of the
emitted electrons to the frequency of
the incident photons and the work
function of the cathode material:
K max = hf − φ
Substitute numerical values for the
light of wavelength λ:
1.20 eV =
Substitute numerical values for the
light of wavelength λ′:
1.76 eV =
Solve these equations simultaneously
for φ to obtain:
φ = 1.04 eV
Substitute in either of the equations
and solve for λ:
λ = 554 nm
=
hc
λ
−φ
hc
λ
−φ
hc
hc
−φ =
−φ
0.800λ
λ'
70 ••
A simple pendulum has a length equal to 1.0 m and has a bob that has
a mass equal to 0.30 kg. The energy of this oscillator is quantized, and the
allowed values of the energy are given by En = (n + 12 )hf0 , where n is an integer
and f0 is the frequency of the pendulum. (a) Find n if the angular amplitude is
1.0°. (b) Find n such that En+1 exceeds En by 0.010 percent.
Wave-Particle Duality and Quantum Physics
Picture the Problem The diagram
shows the pendulum when its angular
displacement is θ. The energy of the
oscillator is equal to its initial potential
energy mgh = mgL(1 − cosθ). We can
find n by equating this initial energy to
En = (n + 12 )hf 0 and solving for n.
θ
L
L cos θ
m
h
Ug = 0
(a) Express the nth-state energy as a
function of the frequency of the
pendulum:
Because the small-amplitude
frequency of the pendulum is given
1 g
:
by f 0 =
2π L
The energy of the pendulum is also
equal to its initial gravitational
potential energy:
Equating these expressions gives:
Solving for n gives:
E n = (n + 12 )hf 0
E n = (n + 12 )
h
2π
g
L
En = mgL(1 − cos θ )
mgL(1 − cos θ ) = (n + 12 )
n=
h
2π
2π m g L3 2 (1 − cosθ ) 1
−
2
h
Substitute numerical values and evaluate n:
2π (0.30 kg ) 9.81 m/s 2 (1.0 m ) (1 − cos1.0°) 1
n=
−
2
6.626 × 10 −34 J ⋅ s
32
= 1.357 × 10 30 = 1.4 × 10 30
(b) Because En+1 exceeds En by 0.010
percent:
g
L
E n +1 − E n
= 1.0 × 10 − 4
En
53
54
Chapter 34
Substituting for En+1 and En gives:
(n + 1 + 12 )hf 0 − (n + 12 )hf 0
(n + 12 )hf 0
or
(n + 1 + 12 ) − (n +
Solving for n yields:
1
2
= 1.0 × 10 − 4
) = 1.0 × 10 −4 (n + 12 )
n = 1.0 × 10 4 − 12 = 1.0 × 10 4
71 ••
[SSM] (a) Show that for large n, the fractional difference in energy
between state n and state n + 1 for a particle in a one-dimensional box is given
approximately by (En +1 − En )/ E n ≈ 2 / n (b) What is the approximate percentage
energy difference between the states n1 = 1000 and n2 = 1001? (c) Comment on
how this result is related to Bohr’s correspondence principle.
Picture the Problem We can use the fact that the energy of the nth state is related
to the energy of the ground state according to En = n 2 E1 to express the fractional
change in energy in terms of n and then examine this ratio as n grows without
bound.
(a) Express the ratio
(En + 1 − En)/En:
En+1 − En (n + 1) − n 2 2n + 1
=
=
En
n2
n2
2
=
2 1
2
+ 2 ≈
n n
n
for n >> 1.
(b) Evaluate
E1001 − E1000
:
E1000
E1001 − E1000
2
≈
= 0. 2%
1000
E1000
(c) Classically, the energy is continuous. For very large values of n, the energy
difference between adjacent levels is infinitesimal.
72 ••
A mode-locked, titanium–sapphire laser has a wavelength of 850 nm
and produces 100 million pulses of light each second. Each pulse has a duration
of 125 femtoseconds (1 fs = 10–15 s) and consists of 5 × 109 photons. What is the
average power produced by the laser?
Picture the Problem We can apply the definition of power in conjunction with
the de Broglie equation for the energy of a photon to derive an expression for the
average power produced by the laser.
Wave-Particle Duality and Quantum Physics
55
ΔE
Δt
The average power produced by the
laser is:
P=
Use the de Broglie equation to
express the energy of the emitted
photons:
ΔE = Nhf =
Nhc
λ
where N is number of photons in each
pulse.
Substitute for ΔE to obtain:
P=
Nhc
λ Δt
Substitute numerical values and evaluate P:
P=
(5 ×10 )(6.626 ×10
9
J ⋅ s )(2.998 × 108 m/s )
≈ 0.1 W
(850 nm )(10 −8 s )
−34
Remarks: Note that the pulse length has no bearing on the solution.
73 ••
This problem estimates the time lag in the photoelectric effect that is
expected classically but not observed. Let the intensity of the incident radiation
falling on an atom be 0.010 W/m2. (a) If the area presented by the atom is
0.010 nm2, find the energy per second falling on an atom. (b) If the work function
is 2.0 eV, how long would it take for this much energy to fall on the atom if the
radiation energy was distributed uniformly rather than in compact packets
(photons)?
Picture the Problem We can find the rate at which energy is delivered to the
atom using the definitions of power and intensity. We can also use the definition
of power to determine how much time is required for an amount of energy equal
to the work function to fall on one atom.
ΔE
= IA
Δt
(a) Relate the energy per second
(power) falling on an atom to the
intensity of the incident radiation:
P=
Substitute numerical values and
evaluate P:
P = 0.010 W/m 2 0.010 × 10 −18 m 2
(
)(
J
1eV
= 10 −22 ×
s 1.602 × 10 −19 J
= 6.242 × 10 −4 eV/s
= 6.2 × 10 −4 eV/s
)
56
Chapter 34
(b) Classically:
Δt =
ΔE φ
=
P
P
Substitute numerical values and
evaluate Δt:
Δt =
2.0 eV
= 3204 s
6.242 ×10 −4 eV/s
≈ 53 min