, SOLUTIONS MANUAL
CHAPTER 1
1. The energy contained in a volume dV is
U(ν,T )dV = U (ν ,T )r 2 dr sinθ dθdϕ
when the geometry is that shown in the figure. The energy from this source that emerges
through a hole of area dA is
dAcos θ
dE(ν ,T ) = U (ν ,T )dV
4πr 2
The total energy emitted is
π /2 2π dA
dE (ν ,T ) = ∫ dr∫ dθ ∫ dϕU (ν,T )sin θ cosθ
cΔt
0 0 0 4π
dA π/2
= 2π cΔtU (ν ,T ) ∫ dθ sinθ cosθ
. 4π 0
1
= cΔtdAU (ν ,T )
4
By definition of the emissivity, this is equal to EΔtdA . Hence
c
E (ν, T ) = U (ν, T )
4
2. We have
c c 8π hc 1
w(λ ,T ) = U (ν ,T ) | dν / dλ |= U ( ) 2 = 5
λ λ λ e hc/λkT
−1
This density will be maximal when dw(λ ,T ) / dλ = 0 . What we need is
A /λ
d ⎛1 1 ⎞ 1 1 e A 1
= (−5 − (− )) =0
d λ ⎝ λ e − 1⎠
5 A / λ
λ λ e −1 λ e λ −1
6 5 A / λ 2 A /
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
, hc
λmaxT = = 2.898 × 10 −3 m
4.965k
In example 1.1 we were given an estimate of the sun’s surface temperature
as 6000 K. From this we get
28.98 × 10 −4 mK
λsun
max = = 4.83 × 10−7 m = 483nm
6 × 10 3 K
3. The relationship is
hν = K + W
where K is the electron kinetic energy and W is the work function. Here
hc(6.626 × 10−34 J .s)(3× 10 8 m / s)
hν = = = 5.68 × 10 −19 J = 3.55eV
λ 350 × 10 m −9
With K = 1.60 eV, we get W = 1.95 eV
4. We use
hc hc
− = K1 − K2
λ1 λ2
since W cancels. From ;this we get
1 λ1λ 2
h= (K − K 2 ) =
c λ 2 − λ1 1
(200 × 10 −9 m)(258 × 10−9 m)
= × (2.3 − 0.9)eV × (1.60 × 10 −19 )J / eV
(3 × 10 8 m / s)(58 × 10−9 m)
−34
= 6.64 × 10 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 backward-
scattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its
recoil momentum is obtained from E = p 2c 2 + m 2c 4 . The energy conservation
equation reads
hν + mc 2 = hν ' + E
and the momentum conservation equation reads
hν hν '
=− +p
c c
, that is
hν = −hν '+ pc
We get E + pc − mc 2 = 2hν from which it follows that
p2 c 2 + m2 c 4 = (2hν − pc + mc 2 ) 2
so that
4 h 2ν 2 + 4 hνmc 2
pc =
4 hν + 2mc 2
The energy loss for the photon is the kinetic energy of the proton
K = E − mc 2 . Now hν = 100 MeV and mc 2 = 938 MeV, so that
pc = 182MeV
and
E − mc 2 = K = 17.6 MeV
6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing
electron momentum. Energy conservation reads
hν + mc = hν ' + p c + m c
2 2 2 2 4
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 ') + (ipx c + jpy c)
Hence px c = hν ; pyc = − hν '. We use this to rewrite the energy conservation equation as
follows:
(hν + mc 2 − hν ')2 = m 2c 4 + c 2 ( px2 + py2 ) = m2 c 4 + (hν ) 2 + (hν ') 2
From this we get
⎛ mc 2 ⎞
hν'= hν ⎜ ⎟
⎝ hν + mc 2 ⎠
We may use this to calculate the kinetic energy of the electron
CHAPTER 1
1. The energy contained in a volume dV is
U(ν,T )dV = U (ν ,T )r 2 dr sinθ dθdϕ
when the geometry is that shown in the figure. The energy from this source that emerges
through a hole of area dA is
dAcos θ
dE(ν ,T ) = U (ν ,T )dV
4πr 2
The total energy emitted is
π /2 2π dA
dE (ν ,T ) = ∫ dr∫ dθ ∫ dϕU (ν,T )sin θ cosθ
cΔt
0 0 0 4π
dA π/2
= 2π cΔtU (ν ,T ) ∫ dθ sinθ cosθ
. 4π 0
1
= cΔtdAU (ν ,T )
4
By definition of the emissivity, this is equal to EΔtdA . Hence
c
E (ν, T ) = U (ν, T )
4
2. We have
c c 8π hc 1
w(λ ,T ) = U (ν ,T ) | dν / dλ |= U ( ) 2 = 5
λ λ λ e hc/λkT
−1
This density will be maximal when dw(λ ,T ) / dλ = 0 . What we need is
A /λ
d ⎛1 1 ⎞ 1 1 e A 1
= (−5 − (− )) =0
d λ ⎝ λ e − 1⎠
5 A / λ
λ λ e −1 λ e λ −1
6 5 A / λ 2 A /
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
, hc
λmaxT = = 2.898 × 10 −3 m
4.965k
In example 1.1 we were given an estimate of the sun’s surface temperature
as 6000 K. From this we get
28.98 × 10 −4 mK
λsun
max = = 4.83 × 10−7 m = 483nm
6 × 10 3 K
3. The relationship is
hν = K + W
where K is the electron kinetic energy and W is the work function. Here
hc(6.626 × 10−34 J .s)(3× 10 8 m / s)
hν = = = 5.68 × 10 −19 J = 3.55eV
λ 350 × 10 m −9
With K = 1.60 eV, we get W = 1.95 eV
4. We use
hc hc
− = K1 − K2
λ1 λ2
since W cancels. From ;this we get
1 λ1λ 2
h= (K − K 2 ) =
c λ 2 − λ1 1
(200 × 10 −9 m)(258 × 10−9 m)
= × (2.3 − 0.9)eV × (1.60 × 10 −19 )J / eV
(3 × 10 8 m / s)(58 × 10−9 m)
−34
= 6.64 × 10 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 backward-
scattered photon energy be hν' . Let the energy of the recoiling proton be E. Then its
recoil momentum is obtained from E = p 2c 2 + m 2c 4 . The energy conservation
equation reads
hν + mc 2 = hν ' + E
and the momentum conservation equation reads
hν hν '
=− +p
c c
, that is
hν = −hν '+ pc
We get E + pc − mc 2 = 2hν from which it follows that
p2 c 2 + m2 c 4 = (2hν − pc + mc 2 ) 2
so that
4 h 2ν 2 + 4 hνmc 2
pc =
4 hν + 2mc 2
The energy loss for the photon is the kinetic energy of the proton
K = E − mc 2 . Now hν = 100 MeV and mc 2 = 938 MeV, so that
pc = 182MeV
and
E − mc 2 = K = 17.6 MeV
6. Let hν be the incident photon energy, hν' the final photon energy and p the outgoing
electron momentum. Energy conservation reads
hν + mc = hν ' + p c + m c
2 2 2 2 4
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 ') + (ipx c + jpy c)
Hence px c = hν ; pyc = − hν '. We use this to rewrite the energy conservation equation as
follows:
(hν + mc 2 − hν ')2 = m 2c 4 + c 2 ( px2 + py2 ) = m2 c 4 + (hν ) 2 + (hν ') 2
From this we get
⎛ mc 2 ⎞
hν'= hν ⎜ ⎟
⎝ hν + mc 2 ⎠
We may use this to calculate the kinetic energy of the electron
