TEST BANK FOR Quantum Physics 3rd Edition By Stephen Gasiorowicz

1. The energy contained in a volume dV is
U(ν,T)dV = U(ν ,T)r2drsinθ dθdϕ
when the geometry is that shown in the figure. The energy from this source that emerges
through a hole of area dA is
dE(ν ,T) = U(ν ,T)dV
dAcosθ
4πr2
The total energy emitted is
.
dE(ν ,T) = dr dθ dϕU(ν,T)sinθ cosθ dA
0 4π
2π ∫ 0
π /2 ∫ 0
cΔt ∫
= dA
4π
2π cΔtU(ν ,T) dθ sinθ cosθ
0
π / 2 ∫
= 1
4
cΔtdAU (ν ,T)
By definition of the emissivity, this is equal to EΔtdA . Hence
E(ν,T) = c
4
U(ν,T)
2. We have
w(λ ,T) = U(ν ,T) | dν / dλ |= U(
c
λ )
c
λ2 = 8πhc
λ 5
1
ehc/λkT −1
This density will be maximal when dw(λ ,T) / dλ = 0. What we need is
d
dλ
1
λ 5
1
eA /λ −1
⎛
⎝
⎞
⎠ = (−5
1
λ6 − 1
λ5
eA /λ
eA/λ −1
(− A
λ 2 ))
1
eA /λ −1
= 0
Where A = hc / kT . The above implies that with x = A /λ , we must have
5 − x = 5e−x
A solution of this is x = 4.965 so that
λ maxT = hc
4.965k
= 2.898 ×10−3m
In example 1.1 we were given an estimate of the sun’s surface temperature
as 6000 K. From this we get
λ max
sun = 28.98 ×10−4mK
6 ×103K
= 4.83 ×10−7m = 483nm
3. The relationship is
hν = K +W
where K is the electron kinetic energy and W is the work function. Here
hν = hc
λ
= (6.626 ×10−34 J .s)(3×108m / s)
350 ×10−9m
= 5.68 ×10−19J = 3.55eV
With K = 1.60 eV, we get W = 1.95 eV
4. We use
hc
λ1
− hc
λ2
= K1 − K2
since W cancels. From ;this we get
h = 1
c
λ1
λ2
λ2
− λ
1
(K1 − K2) =
= (200 ×10−9m)(258 ×10−9m)
(3×108m / s)(58 ×10−9m)
× (2.3− 0.9)eV × (1.60 ×10−19)J / eV
= 6.64 ×10−34 J .s
5. The maximum energy loss for the photon occurs in a head-on collision, with the
photon scattered backwards. Let the incident photon energy be hν , and the backwardscattered
photon energy be hν' . Let the energy of the recoiling proton be E. Then its
recoil momentum is obtained from E = p2c 2 + m2c 4 . The energy conservation
equation reads
hν + mc2 = hν '+E
and the momentum conservation equation reads
hν
c
= − hν '
c
+ p
that is
hν = −hν '+ pc
We get E + pc − mc2 = 2hν from which it follows that
p2c2 + m2c4 = (2hν − pc + mc2)2
so that
pc = 4h2ν 2 + 4hνmc2
4hν + 2mc2
The energy loss for the photon is the kinetic energy of the proton
K = E − mc2 . Now hν = 100 MeV and mc2 = 938 MeV, so that
pc = 182MeV
and
E − mc2 = K = 17.6MeV
6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing
electron momentum. Energy conservation reads
hν + mc2 = hν '+ p2c2 + m2c4
We write the equation for momentum conservation, assuming that the initial photon
moves in the x –direction and the final photon in the y-direction. When multiplied by c it
read
i(hν ) = j(hν ') + (ipxc + jpyc)
Hence pxc = hν ; pyc = −hν '. We use this to rewrite the energy conservation equation as
follows:
(hν + mc 2 − hν ')2 = m2c 4 + c 2(px
2 + py
2) = m2c4 + (hν )2 + (hν ')2
From this we get
hν'= hν mc2
hν + mc2
⎛
⎝
⎜ ⎞
⎠
⎟
We may use this to calculate the kinetic energy of the electron
K = hν − hν '= hν 1− mc2
hν + mc2
⎛
⎝
⎜ ⎞
⎠
⎟ = hν hν
hν + mc2
= (100keV )2
100keV + 510keV
= 16.4keV
Also
pc = i(100keV ) + j(−83.6keV)
which gives the direction of the recoiling electron.
7. The photon energy is
hν = hc
λ
= (6.63×10−34 J.s)(3 ×108m / s)
3×106 ×10−9m
= 6.63×10−17J
= 6.63×10−17 J
1.60 ×10−19 J / eV
= 4.14 ×10−4 MeV
The momentum conservation for collinear motion (the collision is head on for maximum
energy loss), when squared, reads
hν
c
⎛
⎝
⎞
⎠
2
+ p2 + 2
hν
c
⎛
⎝
⎞
⎠ p η
i = hν '
c
⎛
⎝
⎞
⎠
2
+ p'2 +2
hν '
c
⎛
⎝
⎞
⎠ p' η
f
Here η
i = ±1, with the upper sign corresponding to the photon and the electron moving in
the same/opposite direction, and similarly for η
f . When this is multiplied by c2 we get
(hν )2 + (pc)2 + 2(hν ) pc η
i = (hν ')2 + ( p'c)2 + 2(hν ') p'c η
f
The square of the energy conservation equation, with E expressed in terms of
momentum and mass reads
(hν )2 + (pc)2 + m2c 4 + 2Ehν = (hν ')2 + ( p'c)2 + m2c4 + 2E' hν '
After we cancel the mass terms and subtracting, we get
hν(E − η
i pc) = hν '(E'− η
f p'c)
From this can calculate hν' and rewrite the energy conservation law in the form