Solution Manual to
Quantum Physics
, lOMoARcPSD|58847208
SOLUTION MANUAL
CHAPTER 1
1. The energy contained in a volume dV is
U(ν,T)dV = U(ν,T)r 2drsinθdθdϕ
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 dA4πcosr 2θ
The total energy emitted is
dE
.
tdAU(ν,T)
By definition of the emissivity, this is equal to EΔtdA. Hence
E(ν,T)= c U(ν,T)
4
2. We have
w(λ,T) = U(ν,T) |dν/dλ|= U(λc )λc2 = 8πλhc5ehc/λ1kT −1
This density will be maximal when dw(λ,T) /dλ= 0. What we need is
, lOMoARcPSD|58847208
ddλ⎛⎝ λ 15 eA /λ1−1⎞ ⎠ = (−5 16 − λ15 eAe/λA /λ−1(−λA2 )) eA /λ1 −1 = 0
λ
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
λmaxT = hc= 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
λsunmax = 28.986××10103−K4 mK = 4.83×10−7 m = 483nm
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)(3×108 m / s) −19 = 3.55eV
hν= λ = 350×10−9 m = 5.68×10 J
With K = 1.60 eV, we get W = 1.95 eV
4. We use
hc hc
λ1 − λ2 = K1 − K2
since W cancels. From ;this we get
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h = 1λλ2 1−λ2λ1 (K1 −
K2) = c
= (200(3××10108 m−9 /ms)(258)(58××1010−9−m9 m)) × (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 = p 2c 2 + m 2c 4 . The energy
conservation equation reads
hν+ mc = hν'+E
2
and the momentum conservation equation reads
hν= − hν' + p
c c
that is
hν= −hν'+ pc
We get E + pc − mc2 = 2hν from which it follows that
p2c2 + m2c4 = (2hν− pc + mc2)2
so that
4h2ν2 + 4hνmc2
pc = 4hν+ 2mc2
The energy loss for the photon is the kinetic energy of the proton
K = E − mc . Nowhν = 100 MeV and mc = 938 MeV, so that
2 2
Quantum Physics
, lOMoARcPSD|58847208
SOLUTION MANUAL
CHAPTER 1
1. The energy contained in a volume dV is
U(ν,T)dV = U(ν,T)r 2drsinθdθdϕ
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 dA4πcosr 2θ
The total energy emitted is
dE
.
tdAU(ν,T)
By definition of the emissivity, this is equal to EΔtdA. Hence
E(ν,T)= c U(ν,T)
4
2. We have
w(λ,T) = U(ν,T) |dν/dλ|= U(λc )λc2 = 8πλhc5ehc/λ1kT −1
This density will be maximal when dw(λ,T) /dλ= 0. What we need is
, lOMoARcPSD|58847208
ddλ⎛⎝ λ 15 eA /λ1−1⎞ ⎠ = (−5 16 − λ15 eAe/λA /λ−1(−λA2 )) eA /λ1 −1 = 0
λ
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
λmaxT = hc= 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
λsunmax = 28.986××10103−K4 mK = 4.83×10−7 m = 483nm
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)(3×108 m / s) −19 = 3.55eV
hν= λ = 350×10−9 m = 5.68×10 J
With K = 1.60 eV, we get W = 1.95 eV
4. We use
hc hc
λ1 − λ2 = K1 − K2
since W cancels. From ;this we get
, lOMoARcPSD|58847208
h = 1λλ2 1−λ2λ1 (K1 −
K2) = c
= (200(3××10108 m−9 /ms)(258)(58××1010−9−m9 m)) × (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 = p 2c 2 + m 2c 4 . The energy
conservation equation reads
hν+ mc = hν'+E
2
and the momentum conservation equation reads
hν= − hν' + p
c c
that is
hν= −hν'+ pc
We get E + pc − mc2 = 2hν from which it follows that
p2c2 + m2c4 = (2hν− pc + mc2)2
so that
4h2ν2 + 4hνmc2
pc = 4hν+ 2mc2
The energy loss for the photon is the kinetic energy of the proton
K = E − mc . Nowhν = 100 MeV and mc = 938 MeV, so that
2 2
