### Solutions

```PHYS309 Homework 7 solutions
Problems from Krane Chapter 6:
5. (a) The impact parameter, b, and scattering angle, θ, are related by
b=k
zZe2
θ
cot ,
2K
2
where K is the kinetic energy of the incident particle. For scattering of α particles of kinetic energy 5.00
MeV off gold nuclei at 90°, we get
b=
2 × 79 × 1.44 MeV ⋅ fm
= 22.8 fm.
2 × 5.00 MeV
(b) The quantity
w=k
zZe2
2 Kb
is equal to 1 for scattering at 90°. The distance of closest approach is
)
(
rc = b w + w2 + 1 = 54.9 fm.
(c) The ratio of potential energy to kinetic energy at closest approach is
U
= 2 w w + w2 + 1 = 4.83.
K
)
(
Since the total energy is 5.00 MeV, we find that the kinetic and potential energies at closest approach are
0.86 and 4.14 MeV respectively.
10. Inspection of the Rutherford scattering formula shows that the ratio of scattered alphas from two foils
of different elements but the same thickness is
n1Z12
,
n2 Z 2 2
where n1 and n2 are the number densities and Z1 and Z2 the atomic numbers of the scatterers. The mass
densities of silver and gold are 10.49 and 19.30 g cm-3. Their atomic weights are 107.9 and 197.0. Hence
the number densities are essentially the same. The ratio of scattering off silver to gold is (47/79)2 = 0.35.
12. Using primes to denote quantities after the collision, conservation of energy gives
Eα + me c 2 = Eα′ + Ee′ ,
and conservation of momentum gives
pα = pα′ + pe′ .
To start, assume that the kinetic energy of the electron after the collision is small compared to its rest
mass. We can then use the non-relativistic limit of the energy conservation equation:
1 pα 2 1 pα′ 2 1 pe′ 2
=
+
.
2 mα 2 mα 2 me
Eliminating the electron momentum, we get
2
pα 2 − pα′ 2 ( pα − pα′ )
=
.
mα
me
This leads to
me
( pα + pα′ ) = pα − pα′ ,
mα
which shows that the change in momentum of the alpha particle is small. The change in K.E. is
∆K =
pα 2 − pα′ 2
pα − pα′
me 1 pα 2
me
≈ pα
≈4
=4
K.
2mα
mα
mα 2 mα
mα
For K = 8.0 MeV, the kinetic energy gained by the electron is about 4 keV, which is negligible compared
to the kinetic energy of the alpha particle. Also, since the electron kinetic energy is much less than the
electron rest energy, the non-relativistic approximation is appropriate.
17. The Pfund series corresponds to n0 = 5. The longest wavelength occurs for n = n0 + 1 = 6, and the
series limit occurs when n goes to infinity. Hence
2
λlimit 5−2 − 6−2
5
=
= 1−   .
−2
λlongest
5
6
This gives the series limit to be at wavelength 2279 nm.
18. The radius of the Bohr orbit is proportional to the square of the principal quantum number. The
dynamics of the orbit gives v 2 ∝ r −1. Hence the velocity is inversely proportional to the principal
quantum number. The ratio of the velocity of electron in the ground state to the speed of light in vacuum
is equal to the fine structure constant, α. For the n = 3 state, the electron velocity is αc/3 = 7.30 105 m s-1.
The energy levels are inversely proportional to the square of the principal quantum number. The
ionization potential of hydrogen is 13.5984 eV. This is how much energy is need to remove an electron in
the ground state from the atom. Hence the energy of the n = 3 state is -13.5984/9 eV= -1.511 eV. Since U
= 2E, the potential and kinetic energies of the n = 3 state are -3.022 and 1,511 eV, respectively.
23. The energy of the photons is Eγ = hc λ = 21.0 eV. Hence the photo-electrons have energy 21.0 –
13.6 eV = 7.4 eV.
26. The Rydberg parameter for hydrogen is
µH e4
RH = k
.
4π cℏ3
2
(0.1)
where µ H is the reduced mass. For single electrons orbiting a nucleus of charge Ze, the corresponding
expression is
R = k2
µ Z 2e4
.
4π cℏ3
(0.2)
Since the reduced mass is very close to the electron mass for all single electron ions, we see that all
wavelengths of the Lyman series of singly ionized helium are ¼ the corresponding wavelengths of the
hydrogen Lyman series. For He+, the longest wavelength is 30.4 nm and the shortest wavelength is 22.8
nm.
28. The change from the electrostatic force to the gravitational force requires that ke2 be replaced by
GmM, where M is the proton mass and G is Newton’s constant. The ratio of force strengths is
rge =
GmM
= 4 × 10−40.
2
ke
The Bohr radius is now
1 ℏ2
a0 =
= 1.2 × 1029 m.
2
rge kme
(0.3)
This is larger than the Universe!
37. The energy of the first excited state is 3.4 eV. For a lifetime of 10-8 s the energy uncertainty is
∆E ∼
ℏ
τ
=
1 4 × 10−15 eV ⋅ s
= 6 ×10 −8 eV. This energy uncertainty is small compared to the energy of
−8
2π
10 s
the state.
41. (a) Assuming that the nucleus is fixed, the Rydberg constant for the muonic hydrogen atom is
Rmuonic =
mµ
R∞ = 2.3 ×109 m −1.
me
(0.4)
The wavelengths of spectral lines from the muonic atom are 207 times smaller than the wavelengths of
the electronic atom. The shortest wavelength of the muonic Lyman series is 0.44 nm, which is the X-ray
part of the electromagnetic spectrum.
(b) The mass of the muon is 106 MeV, which is 1/9 the mass of the proton. Hence we should correct for
the finite proton mass. The reduced mass is
µ=
mµ m p
= 0.899mµ = 186me .
mµ + m p
The shortest wavelength is 0.49 nm, a change of 10%.
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