HW#2 Solutions.
P 4. From conservation of energy:
hc
= W + eV1
(1)
1
hc
2
= W + eV2
(2)
By subtracting the two equations we can eliminate work function W:
hc
1
hc
2
= eV1 eV2
(3)
Solving for 1/2:
1
=
2
1
1
e
(V1 V2 )
hc
Thus 2 is:
2 =
1
1
1
=
e
(V1 V2 )
h
HW#3 Solutions.
P3.
a) momentum
electron: p=h/=6.6310-34Js/210-10m=3.310-24 kg(m/s2) m s / m
=3.310-24 kg (m/s)
photon: p= h/=3.310-24 kg (m/s) same as electron
b) total energy, using momentum from part (a)
p 2c 2 + m0 c 4 = 8.18771 1014 J
electron: E =
2
P5. (a) Assuming the surface temperature of the sun to be 5700 oK, use Stefans law to
determine the rest mass lost per second to radiation by the sun. Take the suns diameter
to be 1.4 109 m. (b) What fraction of the suns rest mass is lost each year from
e
HW#4 Solutions
P6.
a) In a head-on collision,
1 zZe 2
From eq. 4-4, D =
4 0 (M )
=
(where z = 2 for particle and Z = 79 for gold)
(2)(79)(1.602 1019 C ) 2
= 4.292 1014 m
6
19
40 (5.30 10 eV .1.602 10 / eV )
1
2b
b) From eq 4-5, cot =
2 D
D 4.292 1014 m 6
HW#5 Solutions
P3. a) The wavefunction from example 5-3:
( x, t ) = ( x ) ( t ) = ( x )e
E 2E 1 c
=
=
2m
h
h
1
1c
Substituting E = h gives: 2 =
=
4
2m
So ( t ) = e
b) E = h =
1c
t
2m
h
4
iEt
h
and
c
,
m
c
m
c
1
E= h
2m
1
2E
2
2 h c 4
=
E =
c) Classical l
HW#6
P 9.
(a) The probability of penetrating a barrier is proportional to the transmission coefficient T.
k I Ia
k I Ia 2
From eq. 6-49, T = 1 + e e
E
E
16 1
V0 V0
(
)
As we will see in part (b), kIIa> 1. Thus T~[(expcfw_kIIa)2]-1~expcfw_-2kIIa.
The
HW#8
P2.
a) Want to evaluate l L for a Bohr-Sommerfield elliptical orbit.
From eqn 8-2, l = iA but i = e T where T is the period of revolution.
d
2
dA
From hint: L = mr 2
and d = 2 dA so L = 2 m
. According to Keplers Law an
dt
r
dt
dA A
elliptical orbit
HW #
Solutions
P3.
From eqn. 7-22, E n =
Z 2e 4
2
(4 o ) 2 h 2 n 2
M
m .
m + M
=
For deuterium:
m
2m
For hydrogen:
4m
p
H =
m + m me , Z H = 1
e
p
p
m ,Z = 1
D =
m + 2m e D
p
e
p
He =
m + 4 m me , Z He = 2
e
p
For Helium:
E H : E D : E He = H :