HW 3 SolutionsChapter 4 Structure of the Atom
Chapter 4 1. With more than one electron we are almost forced into some kind of Bohr-like orbits. This
was the dilemma faced by physicists in the early 20th Century.
2. Non-relativistically K = 1 2mv2 and
v =
2K
m =
= 6.4281× 10−2c
Relativistically K = (γ − 1) mc2 so γ = 1 + K/mc2 and
γ = 1 + 7.7 MeV
1− 1 γ2
= 6.4181× 10−2c
The difference is about 10−4c or about 0.16% of the velocity.
3. Conserving momentum and energy:
Mαvα = Mαv α + mev
α = vα −
Mαv2 α = Mα
e
1 +
me
Mα
= 2vα
4. P (θ) = exp
Therefore multiple scattering does not provide an adequate explanation.
* 5. a) With Z1 = 2, Z2 = 79, and θ = 1 we have
b = Z1Z2e2

2 (7.7× 106 eV) cot (0.5) = 1.69× 10−12 m
b) For θ = 90

2 (7.7× 106 eV) cot (45) = 1.48× 10−14 m
432.
* 6.

For the two different angles everything is the same except the angles, so
f(1) f(2)
= 4.00
7. The fraction f is proportional to nt and to Z2 from Equation (4.12). The question states, however, the number of scattering nuclei per unit area is equal so nt is the same for either target. Therefore
N(Au) N(Al)
132 = 36.93
8. From Example 4.2 we know n = 5.90× 1028 m−3. Thus
f = πnt
10−8 m

= 3.49× 10−4
9. a) With all other parameters equal the number depends only on the scattering angles, so
f(90) f(50)
= 0.217
so the number scattered through angles greater than 90 is (10000) (0.217) = 2170.
b) Similarly f(70) f(50)
= cot2 (35) cot2 (25)
= 0.3088
The numbers for the two angles are thus 4435 and 3088 and the number scattered between 70 and 80 is 4435− 3088 = 1347.
* 10. From the Rutherford scattering result, the number detected through a small angle is inversely
proportional to sin4
sin4 (25) = 2.35× 10−4
and if they count 2000 at 6 the number counted at 50 is (2000) 2.35× 10−4
= 0.47 which
is insufficient.
Chapter 4 Structure of the Atom
11. In each case all the kinetic energy is changed to potential energy:
K = −V = |V | = Z1Z2e2
4π0 (r1 + r2)
where r1 and r2 are the radii of the particular particles (as the problem indicates the particles just touch).
a) Z1 = 2 and Z2 = 13 for Al, Z2 = 79 for Au
Al: K = (2)(13)

2.6× 10−15 m + 3.6× 10−15 m = 6.04 MeV
Au: K = (2)(79)

2.6× 10−15 m + 7.0× 10−15 m = 23.7 MeV
b) Now Z1 = 1 and for the two different values of Z2
Al: K = (1)(13)

1.3× 10−15 m + 3.6× 10−15 m = 3.82 MeV
Au: K = (1)(79)

1.3× 10−15 m + 7.0× 10−15 m = 13.7 MeV
12. a) The maximum Coulomb force is at the surface and equal to 2Z2e2/4π0R2. Then
p = Ft = 2Z2e2
(8 MeV) (0.1 nm) = 2.84× 10−4 rad = 0.016
13. a)
(10 MeV) (0.1 nm) = 2.28× 10−4 rad = 0.013
b) These results are comparable in magnitude with those obtained by electron scattering (Example 4.1).
14. a)
v = e√
4π0mr =
c = 1.53c
b)
2 (1.2× 10−6 nm) = −600 keV
c) Clearly (a) is not allowed and (b) is too much energy.
454.
HW 3
Chapter 4 Structure of the Atom
18. The total energy of the atom is −e2/ (8π0r). Differentiating with respect to time:
dE
dt =
e2
Equating this result with the given equation from electromagnetic theory
e2
In a circular orbit d2r
dt2 is just the centripetal acceleration, which is also given by
a = F
dt = − (4π0)2 3m2c3
4e4 a3
Inserting numerical values we find t = 1.55× 10−11 s.
19.
f(Kα) + f(Lα) = f(Kβ)
20. As in Problem 16, v = 2.19× 106 m/s and
L = mvr = 9.11× 10−31 kg
2.19× 106 m/s
Notice that L = h.
Chapter 4 Structure of the Atom
18. The total energy of the atom is −e2/ (8π0r). Differentiating with respect to time:
dE
dt =
e2
Equating this result with the given equation from electromagnetic theory
e2
In a circular orbit d2r
dt2 is just the centripetal acceleration, which is also given by
a = F
dt = − (4π0)2 3m2c3
4e4 a3
Inserting numerical values we find t = 1.55× 10−11 s.
19.
f(Kα) + f(Lα) = f(Kβ)
20. As in Problem 16, v = 2.19× 106 m/s and
L = mvr = 9.11× 10−31 kg
2.19× 106 m/s
Notice that L = h.
Chapter 4 Structure of the Atom
18. The total energy of the atom is −e2/ (8π0r). Differentiating with respect to time:
dE
dt =
e2
Equating this result with the given equation from electromagnetic theory
e2
In a circular orbit d2r
dt2 is just the centripetal acceleration, which is also given by
a = F
dt = − (4π0)2 3m2c3
4e4 a3
Inserting numerical values we find t = 1.55× 10−11 s.
19.
f(Kα) + f(Lα) = f(Kβ)
20. As in Problem 16, v = 2.19× 106 m/s and
L = mvr = 9.11× 10−31 kg
2.19× 106 m/s
Notice that L = h.
21. hc =
e2
m = 1.4400 eV · nm
2
E0 = e2
8π0a0 = −
2
8π (8.8541878× 10−12 F/m) (5.2917725× 10−11 m) = 2.179874× 10−18 J = 13.606 eV
* 22. From Equation (4.31) vn = (1/n)(h/ma0)
n = 1: v1 = 1 1
1.055× 10−34 J · s (9.11× 10−31 kg) (5.29× 10−11 m)
= 2.19× 106 m/s = 0.0073c
n = 2: v2 = 1 2
1.055× 10−34 J · s (9.11× 10−31 kg) (5.29× 10−11 m)
= 1.09× 106 m/s = 0.0036c
n = 3: v3 = 1 3
1.055× 10−34 J · s (9.11× 10−31 kg) (5.29× 10−11 m)
= 7.30× 105 m/s = 0.0024c
E = hc
= 2.86 eV
This is the energy difference between the two states in hydrogen. From Figure 4.16, we see E3 = −1.51 eV so the initial state must be n = 2. We notice that this energy difference exists between n = 2 (with E2 = −3.40 eV) and n = 5 (with E5 = −0.54 eV).
24. The photon energy is
E = hc
= 13.05 eV
This can only be a transition to n = 1 (E1 = −13.6 eV) and because of the energy difference it comes from n = 5 with E5 = −0.54 eV.
25. In general the ground state energy is Z2E0.
a) E = 12E0 = 13.6 eV
The reduced mass does not change this result to three significant digits.
b) E = 22E0 = 54.4 eV
c) E = 42E0 = 218 eV
48
7.
HW 3
Chapter 4 Structure of the Atom
29. From Problem 22 the speed in the n = 3 state is v = 7.30× 105 m/s. The radius of the orbit is n2a0 = 9a0. Then from kinematics
number of revolutions = vt
2π (9) (5.29× 10−11 m) = 2.44× 106
* 30. The energy of each photon is hc/λ = 12.4 eV. Looking at the energy difference between levels in hydrogen we see that E2 − E1 = 10.2 eV, E3 − E1 = 12.1 eV, and E4 − E1 = 12.8 eV. There is enough energy to excite only to the second or third level. In theory it is possible for a second photon to come along and take the atom from one of these excited states to a higher one, but this is unlikely, because the n = 2 and n = 3 states are short-lived.
* 31. We must use the reduced mass for the muon:
µ = mM
m + M =
106 MeV/c2
938 MeV/c2
a)
2
3.00× 108 m/s
b)

c) First series:
= 0.49 nm
= 1.96 nm
= 4.40 nm
32. The reduced mass for this system is µ = mm/(m + m) = m/2 where m is the mass of each particle. Then
r = 4πε0h
8πε0 (2a0) = −E0
2 = −6.8 eV
33. a) As shown in Problem 32, the radius of the orbit is 2a0.
b) With E0 = 6.8 eV (see Problem 32) we have
E = E2 − E1 = −E0
= 243 nm
34. a)
m
b)
m
c)
m
Note that in each case rm − rn = (m + n)a0 which is valid when m and n differ by 1.
35.

We see that the lines match very well when nu is even for helium but not when it is odd. For
example, when nu = 6, for the ionized helium, then the nu = 3 state for hydrogen will be
very similar. In the same fashion, the nu = 8 for the ionized helium will be very similar to
the nu = 4 for hydrogen. Also, all the “matched” lines differ slightly because of the different
Rydberg constant (which is due to the different reduced masses). The differ by a factor of
RHe/RH = 0.99986/0.99946 = 1.0004.
36. As mentioned in the text below Equation (4.38), if each atom can be treated as single-electron
atoms (and the problem states we can make this assumption), then
R = 1
m −1
Using 4 He (M = 4.0026 u), R = 0.999863R∞ = 1.097223× 10
7 m −1
(off by 0.14%)
m −1
m −1
(off by 0.00023%)
37. The derivation of the Rydberg equation is the same as in the text. Because He +
is hydrogen-
like it works with R = Z2RHe and RHe = 1.097223× 10 7
m −1
R = 4 1.097223× 10
λ = c
51
9.
10. n=3 energy is -13.6/42= -1.51 eV so 1.51 eV is needed
Modern Physics 3313 Spring 2010
HW 3