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
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
=
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 pa
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
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
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)
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
. Acc
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