Unformatted text preview: Chapter 28
QUANTUM PHYSICS
[Conceptual questions 2, 3, 10 in Ch 28]
2. According to the uncertainty principle, it is impossible to know a particle’s position and momentum with
unlimited precision at the same time. If we knew the electron were at a specific place around the nucleus, for
example, the uncertainty in its momentum would make it impossible to predict where it would be a short time
later. This makes it impossible for the electron to be in a circular orbit around the nucleus or in any other welldefined trajectory for that matter. In the quantum mechanical picture of the H atom, the electron’s state is
described by a wave function that gives only the probability for the electron to be located at a particular place.
Furthermore, the expression for the angular momentum of the electron differs in the quantum mechanical
description from that in the Bohr model. The two pictures are similar in that the most likely radius at which to find
the electron in the quantum picture is identical with the radius of the circular orbits in the Bohr model.
Furthermore, the energies of the electron states are identical in the two models.
3. The idea of all molecular motion ceasing at absolute zero is in disagreement with the uncertainty principle. There
is always at least some uncertainty in the position and momentum of a particle. Therefore, we should instead say
that the motion and vibration of molecules at absolute zero are at a minimum—as small as possible in accordance
with the uncertainty principle.
10. In most situations, the majority of atoms in a gas are in their ground state. Population inversion refers to the
opposite situation, when most of the atoms in a gas are in an excited state. A laser produces light by the process of
stimulated emission. Here, atoms in an excited state are stimulated by incident photons to decay to their ground
state, emitting an identical photon in the process. The process therefore requires a gas of atoms residing in
metastable excited states—a population inversion. Problems: 28.1, 28.5, 28.7
1. Strategy Find the de Broglie wavelength using Eq. (281). Compare the result to the diameter of the hoop.
Solution
h
h
6.626 × 10−34 J ⋅ s
=
= 1.3 × 10−34 m
λ= =
p mv (0.50 kg)(10 m s)
The wavelength is much too small compared to the diameter of the hoop for any appreciable diffraction to
occur―for a diameter of ∼ 1 m, it’s a factor of 10−34 smaller!
5. Strategy The electron is relativistic, so use p = γ mv to find the magnitude of its momentum. Find the de Broglie
wavelength using Eq. (281).
Solution
(6.626 × 10−34 J ⋅ s) 1 − ( 3 5 )
h 1 − v2 c2
h
h
=
=
= 3.23 pm
λ= =
3
p γ mv
mv
(9.109 × 10−31 kg)( 5 × 3.00 × 108 m s)
2 7. Strategy Assume that the electrons are nonrelativistic. To produce the same diffraction pattern, the electrons
must have the same wavelength as the xrays. Find the de Broglie wavelength using Eq. (281) and the
wavelength of the xrays using E = hc λ .
Solution Equate the wavelength of an electron and the wavelength of an xray photon.
h
hc
h
hc
λe =
and λp =
, so
=
.
pe
Ep
pe Ep
Now substitute pe = me v and solve for the speed. Ep
h
hc
=
, so v =
.
me v Ep
me c
Find the kinetic energy.
Ep 2
Ep 2
1
1
(16 × 103 eV)2
K = me v 2 = me
=
=
= 250 eV
2
2
me 2 c 2 2me c 2 2(511× 103 eV)
The kinetic energy is small compared to the rest energy of an election, so the use of the nonrelativistic equations
for momentum and kinetic energy was valid. Problems: 28.17, [28.24], 28.26, 28.27, 28.33
17. Strategy Use the positionmomentum uncertainty principle, Eq. (282).
Solution Find the uncertainty in the position of the basketball.
1
106 h 106 (6.626 × 10−34 J ⋅ s)
, so Δx ≥
Δx Δp ≥
=
=
=
= 1× 10−29 m .
2
2Δp 2 p ⋅10−6 4π mv 4π (0.50 kg)(10 m s)
24. Strategy The singleslit diffraction minima are given by a sin θ = mλ. The edge of the central fringe corresponds
to m = 1. Since the width of the central fringe is small compared to the slitscreen distance, use small angle
approximations. According to conservation of energy, the kinetic energy of the electrons is related to the potential
difference by K = eV. Use the de Broglie wavelength with p = 2 Kme = 2eVme for the electrons.
Solution Since θ ≈ tan θ = x D and θ ≈ sin θ = λ a , we have λ = ax D , where x is
half the width of the central fringe and D is the distance from the slit to the screen.
Find the width of the slit.
ax h
h
λ=
==
, so
Dp
2eVme
a= hD
hD
=
=
xp x 2eVme ( 1.13×10−3 m
2 ) (6.626 × 10−34 J ⋅ s)(1.00 m)
2(38.0 eV)(1.602 × 10−19 J eV)(9.109 × 10−31 kg) x
θ D = 352 nm . 26. Strategy Use the energytime uncertainty principle, Eq. (283).
Solution Find the uncertainty in the gammaray energies.
1
1× 10−34 J ⋅ s
1 eV
ΔE Δt ≥
, so ΔE ≈
=
×
= 3 × 10− 4 eV .
2
2Δt 2(1× 10−12 s) 1.602 × 10−19 J
2
27. Strategy The kinetic energy of the electron is given by K = pn (2m), where pn = nh (2 L). The minimum kinetic
energy is the n = 1 state. Solution Find the minimum kinetic energy.
p 2 n2 h2
12 (6.626 × 10−34 J ⋅ s)2
1 eV
=
×
= 380 GeV
K= n =
−31
−15
2
2
2m 8mL
8(9.109 × 10
kg)(1.0 × 10 m) 1.602 × 10−19 J 33. Strategy The energy of a photon is related to its wavelength by E = hc λ . Use Eqs. (288), (289), and (2810).
Solution
(a) Calculate the wavelength.
hc
hc
hc
hc 1240 eV ⋅ nm
λ=
=
=
=
=
= 10.3 nm
E E2 − E1 22 E1 − E1 3E1
3(40.0 eV)
(b) According to Eq. (288), E ∝ L−2 , so doubling the length of the box would reduce the energy to one fourth as
much as before. Problems: [28.36], 28.37, 28.39, [28.43]
36. Strategy There are 4 + 2 electron states in a subshell. Use Table 28.1.
Solution Since = 1, there are 4(1) + 2 = 6 electron states. For = 1, m = −1, 0, and 1; for each m ,
ms = ± 1 2. The states are given in the table. 3 3 3 3 3 3 1 1 1 1 1 1 m −1 −1 0 0 1 1 ms −1
2 +1
2 −1
2 +1
2 −1
2 +1
2 n 37. Strategy Use Table 28.1 and Eq. (2813).
Solution Since n = 3, can be 0, 1, or 2, and m can have values −2, −1, 0, 1, and 2 (for = 2). Lz = m , so it can have values: −2 , − , 0, , and 2
39. Strategy and Solution Since n = 7, . can have values from 0 to 6. For each value of , there are 2(2 + 1) states. So, for = 0, 1, 2, 3, 4, 5, and 6, there are 2, 6, 10, 14, 18, 22, and 26 electron states, respectively. The total is 98.
43. Strategy Lithium, sodium, and potassium have 3, 11, and 19 electrons in a neutral atom, respectively. None of
these appear in the list of exceptions to the subshell order, so subshells are filled in the order listed in Eq. (2816).
The Periodic Table of the elements is arranged in columns by electron configuration.
Solution
(a) The groundstate electron configurations are the following:
Li: 1s 2 2s1; Na: 1s 2 2 s 2 2 p 6 3s1; K: 1s 2 2s 2 2 p 6 3s 2 3 p 6 4s1 . (b) All three neutral atoms have valence +1. Their outermost electron is in the s1 subshell . This is why they
are placed in the same column. Problems: 28.51, [28.55]
51. Strategy The wavelength of a photon is related to its energy by E = hc λ .
Solution The wavelength for this transition is
hc
1240 eV ⋅ nm
=
= 544 nm .
λ=
ΔE 20.66 eV − 18.38 eV
Light of wavelength 544 nm appears green in color.
55. Strategy Refer to Example 28.6. The ratio of the tunneling currents is equal to the ratio of the tunneling
probabilities.
Solution Compute the ratio of the tunneling currents.
9
−1
−9
I a′
= e−2κ ( a′− a ) = e−2(7.245×10 m )(0.020×10 m) = 0.75
Ia
The percentage change is ⎛I
⎞
I a′ − I a
× 100% = ⎜ a′ − 1⎟ × 100% = (0.75 − 1) × 100% = −25% .
Ia
Ia
⎝
⎠ Problems: [28.58], [28.71]
58. Strategy Use the energytime uncertainty principle. The rest energy of a particle is mc 2 .
Solution
(a) Compute the lifetime of an electron created from the vacuum.
1
6.582 × 10−16 eV ⋅ s
ΔE Δt =
, so Δt =
=
=
= 6.440 × 10−22 s .
2
2ΔE 2me c 2 2(0.5110 × 106 eV)
(b) Compute the lifetime of a shot put created from the vacuum.
1.055 ×10−34 J ⋅ s
Δt =
=
=
= 8 × 10−53 s
2
2ΔE 2mc
2(7 kg)(3.00 × 108 m s ) 2
71. Strategy The energy difference is equal to the energy of the photon. Use Eqs. (289) and (2810) to find the
groundstate energy and the energies of the photons emitted for each of the transitions from the n = 3 state to the
ground state. Use Figure 28.10 to sketch the wave function for the third excited state. How the energy level
spacings depend upon the length of the box is given by Eq. (288) for the groundstate energy.
Solution
(a) Find the groundstate energy.
ΔE = E2 − E1 = 22 E1 − E1 = 3E1 = 0.20 eV, so E1 = 0.067 eV . (b) The third excited state is the n = 4 state. The possible transitions are:
4 → 3 → 2 → 1, 4 → 3 → 1, 4 → 2 → 1, and 4 → 1. Find the energies of the transitions. E4 − E3 = 42 E1 − 32 E1 = 7 E1 E3 − E2 = 32 E1 − 22 E1 = 5 E1 E4 − E1 = 42 E1 − E1 = 15 E1 E2 − E1 = 22 E1 − E1 = 3E1 E4 − E2 = 42 E1 − 22 E1 = 12 E1 E3 − E1 = 32 E1 − E1 = 8E1 By increasing energy, the photon energies are 0.20 eV, 0.33 eV, 0.47 eV, 0.53 eV, 0.80 eV, and 1.0 eV.
(c) The figure shows the wave function for the electron in the third
excited state that is confined to a onedimensional box of length L. ψ 0 L x (d) According to Eq. (288), E ∝ L−2 , so increasing L decreases the energy of each level. Therefore, the energy
level spacings would be smaller. Chapter 29
NUCLEAR PHYSICS
Problems: 29.5, 29.11. 29.14
5. Strategy The nucleon number A is the sum of the total number of protons Z and neutrons N. Use the Periodic
Table of the elements to find the number of protons.
Solution Find the number of protons.
Potassium has atomic number 19, so Z = 19.
N = # of neutrons = 21
Find the nucleon number.
A = Z + N = 19 + 21 = 40
So, the symbol is 40
19 K . 11. Strategy A deuteron has 1 proton and 1 neutron. Use Eqs. (297) and (298) to find the mass defect and binding
energy.
Solution Find the mass defect.
Δm = 1× 1.007 276 5 u + 1× 1.008 664 9 u − 2.013 553 u = 0.002 388 u
The binding energy is EB = (Δm)c 2 = 0.002 388 4 u × 931.494 MeV u = 2.225 MeV .
14. Strategy The nucleon number A is the sum of the total number of protons Z and neutrons N. Use Eqs. (297) and
(298) to find the mass defect and binding energy. The binding energy per nucleon is the binding energy divided
by the total number of nucleons in the nucleus.
Solution The 31
15 P atom has 15 protons and 16 neutrons. Its mass is 30.973 761 5 u. Find the mass defect. 31
Δm = (mass of 15 1H atoms and 16 neutrons) − (mass of 15P atom)
= 15 × 1.007 825 0 u + 16 × 1.008 664 9 u − 30.973 761 5 u = 0.282 251 9 u
Calculate the average binding energy per nucleon.
EB (Δm)c 2 0.282 251 9 u × 931.494 MeV u
=
=
= 8.48116 MeV nucleon
A
A
31 nucleons Problems: 29.21, 29.25, [29.27], 29.33, 29.34, [29.35], [29.38], 29.46, 29.47
21. Strategy In betaminus decay, the atomic number Z increases by 1 while the mass number A remains constant.
Use Eq. (2911).
Solution
For the parent 40
( 19 K ) Z = 19, so the daughter nuclide will have Z = 19 + 1 = 20, which is the element Ca. The symbol for the daughter is 40
20 Ca . 25. Strategy The kinetic energy of the decay products equals the energy associated with the change in mass.
Solution Find the change in mass during the decay using atomic masses.
Δm = (mass of 222 Rn + mass of 4He) − (mass of 226Ra) = (222.017 570 5 u + 4.002 603 2 u) − 226.025 402 6 u
86
2
88
= − 0.005 228 9 u
Find the kinetic energy. K = E = Δm c 2 = 0.005 228 9 u × 931.494 MeV u = 4.8707 MeV
Assuming the 222
86 Rn nucleus takes away an insignificant fraction of the kinetic energy, the alpha particle’s kinetic energy will be 4.8707 MeV.
27. Strategy In Problem 21, the daughter nuclide in this decay was found to be 40 Ca. The maximum kinetic energy
20
of the beta particle is equal to the disintegration energy.
Solution The reaction is
40
19 K → 40Ca + −0 e + 0ν
20
1
0 40
The atomic masses of 19 K and 40 Ca are 39.963 998 7 u and 39.962 591 2 u, respectively. To get the masses of
20
the nuclei, we subtract Zme from each. The mass of the electron is 0.000 548 6 u, and the neutrinos mass is
negligible. The mass difference is
Δm = [(M Ca − 20me ) + me ] − (M K − 19me ) = M Ca − M K = 39.962 591 2 u − 39.963 998 7 u = − 0.001 407 5 u. The disintegration energy is E = Δm c 2 = 0.001 407 5 u × 931.494 MeV u = 1.3111 MeV.
The maximum kinetic energy of the β − particle is 1.3111 MeV .
33. Strategy The activity is reduced by a factor of two for each halflife. Use Eqs. (2918), (2920), and (2922) to
find the initial number of nuclei and the probability of decay per second.
Solution (a) Find the number of halflives.
600.0 s
600.0 s =
= 3.000 halflives
200.0 s halflife
3.000 ⎛1⎞
The activity after 3.000 halflives will be R = ⎜ ⎟
⎝2⎠ × R0 = 1
× 80, 000.0 s −1 = 10, 000 s −1 .
8.000 (b) Find the initial number of nuclei.
1
RT
200.0 s
N 0 = R0 = τ R0 = 0 1/ 2 = 80, 000.0 s −1 ×
= 2.308 × 107
ln 2
ln 2
λ
(c) The probability per second is λ = 1 τ = ln 2
ln 2
=
= 3.466 × 10−3 s −1 .
T1/ 2 200.0 s 34. Strategy The activity as a function of time is given by R = R0e−t τ . Use Eq. (2922) to find the time constant.
Solution Find the number of days for the activity to decrease to 2.5 × 106 Bq.
e −t τ = R
t
R
R
8.0 d
2.5 ×106 Bq
, so − = ln
or t = −τ ln
=−
× ln
= 64 d .
R0
R0
R0
ln 2
τ
6.4 × 108 Bq 38. Strategy The activity is given by R = R0e−t τ = R0e−t ln 2 T1/ 2 .
Solution Solve for the halflife.
R
e−t ln 2 T1 / 2 =
R0
t ln 2
R
−
= ln
T1/ 2
R0
−t ln 2
12 min × ln 2
T1/ 2 =
=−
= 2.4 min
R
2.0×103 Bq
ln R
ln
0 6.4×104 Bq 46. (a) Strategy The absorbed dose of ionizing radiation is the amount of radiation energy absorbed per unit mass of
tissue. The number of photons that must be absorbed is equal to the total energy absorbed divided by the
energy per photon.
Solution Calculate the energy absorbed.
energy (E )
= absorbed dose
mass (m)
E = m(absorbed dose) = 0.30 kg × 2000.0 Gy = 600 J
Calculate the number of photons.
1 eV
600 J ⋅
total energy (E )
1.602×10−19 J
# of photons =
=
= 3.7 × 1016 photons
energy per photon 100.0 × 103 eV photon
(b) Strategy Use Eq. (144).
Solution Find the temperature increase.
Q
600 J
Q = mcΔT , so ΔT =
=
= 0.48°C .
mc 0.30 kg × 4186 J (kg ⋅ K) 47. Strategy The absorbed dose of ionizing radiation is the amount of radiation energy absorbed per unit mass of
tissue. Use Eq. (2928a) to find the biologically equivalent dose in sieverts.
Solution Find the total energy absorbed by the tumor.
E = (1.16 × 1017 protons)(950 × 103 eV proton )(1.602 ×10−19 J eV )
Find the biologically equivalent dose.
(1.16 ×1017 protons)(950 × 103 eV proton )(1.602 ×10−19 J eV )
absorbed dose × QF =
× 3.0 = 1.4 × 1010 Sv
3.82 × 10−6 kg Problems: 29.49, 29.56, [29.54], [29.69]
49. Strategy Write out the reaction using variables for the unknown quantities. Balance the reaction to find the
unknowns. The total charge and total number of nucleons must remain the same. The emission of an electron is
betaminus decay.
Solution
(a) The reaction is
0
a
c
e
1
0 n + b (?1) → d (?2 ) → f (?3 ) + −1e
e
f (?3 ) → 4α + 4α
2
2 Working backward, e = 4 + 4 = 8 and f = 2 + 2 = 4 , so (?3 ) = 8
4 Be . Next, c = e = 8 and d = f − 1 = 4 − 1 = 3 , so (?2 ) = 8 Li . Finally, a + 1 = c = 8, so a = 7 and b = d = 3, which means
3
7
(?1) = 3 Li . (b) Yes; the emission of an electron (betaminus decay) is accompanied by the emission of one antineutrino.
54. Strategy From Figure 29.2, the binding energies per nucleon for 235
139
92 U, 54Xe, 95
and 38Sr are approximately 7.6 MeV, 8.3 MeV, and 8.7 MeV, respectively. The energy released is equal to the increase in the binding energy.
Solution The binding energies are as follows:
235
92 U: 235 × 7.6 MeV = 1786 MeV, 139 Xe: 139 × 8.3 MeV = 1154 MeV, 95Sr: 95 × 8.7 MeV = 827 MeV.
38
54 Find the energy released.
1154 MeV + 827 MeV − 1786 MeV = 195 MeV ≈ 200 MeV 56. (a) Strategy The mass numbers on the two sides of the reaction must be equal. Let x be the number of neutrons.
Solution Find the number of neutrons.
235 + 1 = 141 + 93 + x, so x = 2 . (b) Strategy From Figure 29.2, the binding energies per nucleon of 235 U, 141Cs, and 93Rb are approximately
7.6 MeV, 8.35 MeV, and 8.7 MeV, respectively. The energy released is equal to the increase in the binding
energy.
Solution The binding energies of the three nuclides are as follows:
U: 235 × 7.6 MeV = 1786 MeV, 141Cs: 141× 8.35 MeV = 1177 MeV, 93 Rb: 93 × 8.7 MeV = 809 MeV.
Find the binding energy.
235 1177 MeV + 809 MeV − 1786 MeV = 200 MeV
The energy released is approximately 200 MeV. (c) Strategy The atomic masses of 235
141
92 U, 55Cs, and 93 Rb are 235.043 923 1 u, 140.920 044 0 u, and
37 92.922 032 8 u, respectively. Atomic masses can be used, since both sides include the same number of
electrons (92).
Solution Find the change in mass.
Δm = 140.920 044 0 u + 92.922 032 8 u + 2 × 1.008 664 9 u − 235.043 9231 u − 1.008 664 9 u = − 0.193181 4 u
The energy released is E = Δm c 2 = 0.193 181 4 u × 931.494 MeV u = 179.947 MeV .
(d) Strategy Divide the energy released by the rest energy.
Solution
ΔE
Δm
0.193181 4 u
=
=
≈ 0.000 822
E0
m
235.043 9231 u
69. (a) Strategy The nucleon number A is the sum of the total number of protons Z and neutrons N.
Solution The reaction is
226
88 Ra → 222 Rn + 4α .
86
2
There will be 226 − 4 − 86 = 136 neutrons and 88 − 2 = 86 protons. (b) Strategy The activity is related to the number of nuclei N and the time constant τ by R = N τ . The time
constant is related to the halflife by T1 2 = τ ln 2.
Solution Calculate the activity of the radon.
107 ⋅ ln 2
N N ln 2
=
=
= 20 decays/s
R=
τ
3.8 d × 24 h d × 3600 s h
T1/ 2
So, 20 alpha particles per second are emitted due to the decaying radon nuclei. ...
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This note was uploaded on 09/30/2011 for the course PHYS 1101 at Cornell.
 '08
 RICHARDSON, B
 Physics, Momentum, Quantum Physics

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