Download Relativity Problem Set 7 - Solutions Prof. J. Gerton October 24, 2011

Relativity Problem Set 7 - Solutions
Prof. J. Gerton
October 24, 2011
Problem 1
(10 pts.) Quantization of the Bohr model
In Bohr’s model , the velocity of the electron is quantized as vn = α c/n, where
α = 1/137 is the fine structure constant. Since αc = 2.19 × 106 m/s, we see that
only the velocities in (b) and (d) are allowed, being given by the above relation with
n = 1 and n = 2 respectively. The quantities in (a) and (c) are not associated with
an integer quantum number, so they are not allowed.
Problem 2
(10 pts.) Lyman spectrum
An electron de-exciting in a Hydrogen atom from a state of quantum number ni to a
final state with nf = 1 emits a photon of frequency
hν = Eni − E1 .
(1)
Since the energy spectrum of the Hydrogen is
α2 me c2
13.6eV
=−
,
2
2n
n2
(2)
1
α 2 m e c2
1
−1 =
1− 2 .
n2i
2
ni
(3)
En = −
we have
α2 me c2
hν = −
2
Since ν = c/λ, we finally have
1
α 2 m e c2
=
λ
2hc
1
1− 2
ni
.
(4)
The Rydberg constant is thus
R=
α 2 m e c2
.
2hc
1
(5)
2
Problem 3
(10 pts.) Positronium
The electron and the positron are a distance r apart, and they orbit around their
mutual center of mass which is at a distance r/2. Each particle has mass me and
velocity v.
We use the Bohr-Sommerfeld quantization rule for the angular momentum
L = n~,
(6)
where n is an integer, also known as the principal quantum number. Each particle
has angular momentum me vr/2, so the total angular momentum is L = me vr and
rme v = n~.
(7)
Another equation is found by equating the centripetal force Fc and the electromagnetic force Fe . We have
2me v 2
v2
=
,
(8)
Fc = me
r/2
r
and
e2
α ~c
Fe =
= 2 ,
(9)
2
4π0 r
r
where α ≈ 1/137 is called the fine structure constant. Equating Fc = Fe , we find the
radius
α ~c
(10)
r=
2me v 2
Solving Eqs. (7) and (10) for r, we find
r=
2n2 ~c
.
αme c2
(11)
We now find the energy spectrum E. The total energy E is given by the sum of
the kinetic energy and of the potential energy. In a Coulomb potential, E = U/2 (the
so-called virial theorem, valid also in classical mechanics). So,
U
α~c
α2 me c2
E(= En ) =
=−
=−
.
(12)
2
2r
4n2
In the last equation we have used the expression for the radius in Eq. (11). The label
n in En is added so that it is clear that the energy is quantized in this system, thanks
to the quantization of the angular momentum. Plugging in numbers,
6.81 eV
.
(13)
n2
The energy spectrum is proportional to n2 as in the Hydrogen atom case, which is
obvious because we are still treating a Coulomb field, but the number of the fundamental level E0 = 6.81 eV is different from the fundamental level of the Hydrogen
system. Positronium has long being theoretically predicted and described, but only
recently it has been possible to actually create this system.
En = −
course name
PS #
3
Problem 4
(10 pts.) Rutherford scattering
(a) The α particle stops when its initial kinetic energy is equal to the electrostatic
potential energy, so
Qα QAu
K=U =
.
(14)
4π0 D
The potential energy is
Qα QAu
(+2e) (+79e)
=
.
4π0 D
4π0 D
U=
(15)
Using the numerical value
e2
= 1.438 MeV fm,
4π0
we have
U = 1.438 MeV fm
fm
(2) (79)
= 227.2 MeV .
D
D
(16)
(17)
Equating U with K, we obtain
D = 8.4 fm.
(18)
This is an upper limit on the size of the Gold nucleus. In practice, this is larger
than the actual size of such nucleus, that can be estimated as 1.3 fm (A)1/3 ≈ 7.5
fm, where A = 197 for Gold.
(b) We use the equation derived in the lecture, in the form
b=
Qα QAu
1
.
8π0 K tan θ/2
(19)
(+2e) (+79e)
1
= 7.3 fm.
2 (27 MeV) tan 30o
(20)
Plugging in numbers, we have
b = 1.438 MeV fm
Problem 5
(10 pts.) Geiger-Marsden experiment
(a) We use the same reasoning as in Problem 1, part (a). Equating U with K = 7.7
MeV, we find
fm
7.7 MeV = 227.2 MeV ,
(21)
D
from which the upper value for the radius is D = 29.5 fm.
course name
PS #
4
(b) We assume that each Gold nucleus is packed in a sphere of radius r < D, thus of
volume
4π 3
D = 1.075 × 105 fm3 .
(22)
V <
3
(Using a cube to pack up the atom works as fine, for the estimate). Using
u = 1.661 × 10−27 Kg, each Gold nucleus has a mass mAu = 3.271 × 10−25 kg
(we neglect the masses of the electrons). So, the density has a lower value
ρ=
3.271 × 10−25 kg
mAu
15
3
>
3 = 3.043 × 10 kg/m .
5
V
1.075 × 10 fm
(23)
This is the typical order of magnitude for nuclear matter density.
course name
PS #