Download Homework 2 Solutions

Homework 2 Solutions
Problem 1
A)
The charges involved are now −e and Ze so Coulomb’s law becomes
F =
−1 Ze2
r̂
4π0 r2
So, in the Bohr atom everything will be the same except we must make the
replacement e2 → Ze2 . For the orbital radius we get
rn =
4π0 h̄2 2
a0
n = n2
µZe2
Z
and the energy is
En = −
Z2
µZ 2 e4 1
=
−
(13.6 ev)
2
2
2
n2
32π 2 0 h̄ n
So, to summarize, the radius is decreased by a factor of Z and the energy is
increased by a factor of Z 2 .
B) and C)
We just need to plug in 2 and 3 for Z for helium and lithium respectively and
take n = 1. For helm we get
r1 = .27Å
E1 = −54.4 eV
and for lithium we get
r1 = .18Å
E1 = −122 eV
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Problem 2
The energy of the photon is given by the energy loss of the electron, the difference
between the energy levels.
1
1
− 2
E = (13.6 eV)
n22
n1
and the wavelength is then given by equation 5.5
λ=
hc
E
Now, there are 3 possible transitions corresponding to (n1 , n2 ) = (3, 2), (3, 1), (2, 1).
Plugging in gives us
transition E (eV) λ nm
(3 → 2)
1.9
656.5
(3 → 1)
12.1
102.6
(2 → 1)
10.2
121.6
Problem 3
A)
We always have ms = ±1/2. The other quantum numbers are given by
• n=1
– l=0, m=0
• n=2
– l=0, m=0
– l=1 ,m={-1,0,1}
• n=3
– l=0, m=0
– l=1, m={-1,0,1}
– l=2, m={-2,-1,0,1,2}
B)
There are 2 good ways to do this problem, doing the sum explicitly or using
induction.
In the explicit method we write the degeneracy as a sum as follows, using
the fact that there are 2l + 1 choices of m for each l.
gn = 2
n−1
X
(2l + 1)
l=0
2
Then, we can get the answer by using the fact that the sum of the first n
consecutive numbers is given by n(n + 1)/2.
!
n−1
X
gn = 4
l + 2n
l=0
(n − 1)n
=4
+ 2n
2
= 2n2 − 2n + 2n
= 2n2
The proof by induction works by noticing that the states in the (n+1)th
energy level have all of the same orbital angular momentum states in the nth
energy level, plus the additional states that have l = n which are not present in
the nth energy level. Since there are 2(n + 1) states with l = n this means that
the difference in degeneracy between adjacent energy levels is 2(n + 1).
Now, we can follow the standard induction procedure to prove that the
degeneracy is 2n2 . Check that it is true for n = 1, which is trivial, and prove
that it is true for n+1 given that it is true for n. If we assume the degeneracy
of the nth energy level is 2n2 we know that the degeneracy for n+1 is
gn+1 = 2n2 + 2(n + 1)
= 2(n2 + n + 1)
= 2(n + 1)2
and we are done.
Problem 4
Equation 9.7 gives the energy density of a blackbody. Multiplying by the volume
of an eyeball gives the energy
E=
4 3 4
πr aT = 9.9 × 10− 11 J
3
The inverse square law gives the flux from a light bulb at one meter F = L/4πr2 ,
where we will take L to be 100 watts. Multiplying by the area of the pupil gives
the energy entering the eye per second and multiplying this by the time the
light spends in the eye, 2r/c, gives the total energy in the eye at any given time.
E=
2LAr
= 8.0 × 10−15 J
4πr2 c
The reason it is dark when you close your eyes is that the blackbody radiation
from your eye peaks in the infrared, so it is not visible.
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Problem 5
A)
First, we need to determine an expression for the number density. Equation 9.5
gives the energy density per wavelength as
uλ =
8πhc
λ5 (ehc/λkT − 1)
and the energy of a photon of wavelength λ is E = hc/λ. Thus, the number
density of photons at a specific wavelength is
nλ =
8π
λ4 (ehc/λkT
− 1)
If we integrate this expression over all wavelengths we get the total number
density. The answer is
8πk 3 T 3
n = 2.4 3 3
h c
Now, the average energy per photon is given by the energy density divided by
the number density. The energy density is given by u = 4σT 4 /c. Dividing gives
the answer.
B)
Plugging into the above equation yields u/n = 3650 eV at the center of the sun
and 1.34 eV in the photosphere.
Problem 6
The photons you see originate at an optical depth of 2/3. So, we have
d=
2
= 18.5 m
3κ500 ρ
Problem 7
In the Eddington approximation we take the intensity to be a constant Ii in the
−z direction and another constant Io in the +z direction. The average intensity
is given by equation 9.3. Plugging in for I yields
Z 2π Z π
1
hIi =
I sin θ dθdφ
4π 0
0
!
Z π/2
Z π
1
Io
sin θ dθ + Ii
sin θ dθ
=
2
0
π/2
=
1
(Ii + Io )
2
4
Equation 9.8 gives the flux
π/2
Z
F rad = 2π Io
cos θ sin θ dθ + Ii
0
!
π
Z
cos θ sin θ dθ
π/2
= π(Io − Ii )
And equation 9.9 gives the radiation pressure
2π
P rad =
c
Z
π/2
Io
Z
2
cos θ sin θ dθ + Ii
0
!
π
2
cos θ sin θ dθ
π/2
2π
(Io + Ii )
3c
4π
hIi
=
3c
=
Problem 8
For a plane-parallel grey atmosphere in equilibrium we know from equation 9.44
that the mean intensity is equal to the source function. Then, equation 9.50
yields the result.
4π
2
hIi = Frad τ +
3
3
4π
2
S(τ ) = Frad τ +
3
3
πS(2/3) = Frad
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