Solutions Manual for Quantum Physics, 3rd Edition by Stephen Gasiorowicz

, SOLUTIONS MANUAL
CHAPTER 1
1. The energy contained in a volume dV is

U(,T )dV = U (,T )r 2dr sindd

when the geometry is that shown in the figure. The energy from this source that emerges
through a hole of area dA is
dAcos
dE(,T ) = U (,T )dV
4r 2

The total energy emitted is
 /2 dA
 dr d  dU (,T ) sin cos
c t 2
dE(,T ) =
0 0 0 4
dA  /2
= 2ctU(,T )  d sin cos
. 4 0

1
= ctdAU (,T )
4

By definition of the emissivity, this is equal to EtdA . Hence

c
E(,T ) = U (,T )
4

2. We have
w(,T ) = U (,T ) | d / d |= c c 8hc 1
U( ) 2 =
  5 ehc/kT − 1

This density will be maximal when dw(,T ) / d = 0. What we need is
d 1 1  1 1 eA / A 1
= (−5 − (− )) =0
d 5 eA / − 1 6 5 eA/ − 1 2 eA / − 1

Where A = hc / kT . The above implies that with x = A / , we must have

5 − x = 5e−x

A solution of this is x = 4.965 so that

, hc
maxT = = 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
 sun
= = 4.83  10−7 m = 483nm
max
6  103K

3. The relationship is

h = 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)
h = = = 5.68  10−19 J = 3.55eV
 350  10−9 m

With K = 1.60 eV, we get W = 1.95 eV

4. We use
hc hc
− = K1 − K2
1 2

since W cancels. From ;this we get
1
h= 12 (K − K ) =
c 2 − 1
1 2


(200  10−9 m)(258  10−9 m)
=  (2.3− 0.9)eV  (1.60  10 −19 )J / eV
(3 10 m / s)(58  10 m)
8 − 9

= 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 h , and the backward-
scattered photon energy be h' . Let the energy of the recoiling proton be E. Then its
recoil momentum is obtained from E = p2c 2 + m 2c 4 . The energy conservation
equation reads

h + mc2 = h'+ E

and the momentum conservation equation reads
h h'
=− +p
c c

, that is
h = −h'+ pc

We get E + pc − mc2 = 2h from which it follows that

p2c2 + m2c4 = (2h − pc + mc2) 2

so that

4h22 + 4hmc2
pc =
4h + 2mc2

The energy loss for the photon is the kinetic energy of the proton
K = E − mc2 . Now h = 100 MeV and mc 2 = 938 MeV, so that

pc = 182MeV
and

E − mc2 = K = 17.6MeV

6. Let h be the incident photon energy, h' the final photon energy and p the outgoing
electron momentum. Energy conservation reads

h + mc 2 = h'+ 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(h) = j(h') + (ipxc + jpyc)

Hence px c = h; pyc = − h'. We use this to rewrite the energy conservation equation as
follows:
(h + mc 2 − h')2 = m 2c 4 + c 2 (p 2 + p2 ) = m2c4 + (h)2 + (h')2
x y


From this we get

 mc 
2

h'= h
 h + mc 2 

We may use this to calculate the kinetic energy of the electron