PHY 3101
Summer 2007
Exam #1
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1. A heavy version of the electron, called the muon, differs from the
electron only in that its mass is mμ≈207me. (a) What is the radius of the
muonic hydrogen atom in its ground state? (b) Calculate the
wavelength of radiation emitted in the transition of a muonic hydrogen
atom from the n=2 state to the n=1 state.
Solution:
(a) For the radius of the ground state, we
have
n 2 2
1 (1.055 10 34 J  s ) 2
r

km e 2 (8.98 109 N  m 2 / C 2 )( 207  9.111031 kg)(1.6 10 19 C ) 2
 2.6 1013 m
(b) Because the energy levels of hydrogen are proportional to me , the
energy level of the muonic hydrogen atom ground state is:
E0  
m k 2 e 4
2 2
207me k 2 e 4

 207  (13.6eV )  2815eV
2 2
For the state n=2:
ΔE=2111eV;  
E2 
E0
2815eV

 704eV
2
n
4
hc 1240eV  nm

 0.587nm  5.9  10 10 m
E
2111eV
2. X-rays having energy of 300 keV undergo Compton scattering from
a target. The scattered rays are detected at 37.0° relative to the
incident rays. Find (a) the Compton shift at this angle, (b) the energy of
the scattered x-ray, and (c) the energy of the recoiling electron.
Solution:
h



1 cos  :
(a)
m ec
 
(b)
E0 
6.626  1034
9.11 10  3.00  10 
31
8
1 cos37.0 
4.88  1013 m
hc
0 :
 300 10
3

19
eV 1.60  10
J eV

6.626  10  3.00  10


34
8
0
0  4.14  1012 m
   0    4.63 1012 m
and



34
8
hc 6.626  10 J s 3.00  10 m s
E 

 4.30  1014 J 268 keV
12

4.63  10 m
(c)
K e  E0  E  300 keV  268.5 keV  31.5 keV

m s
3. An unstable particle with a mass of 3.34 × 10–27 kg is initially at rest.
The particle decays into two fragments that fly off along the x axis with
velocity components 0.987c and –0.868c. Find the masses of the
fragments. (Suggestion: Conserve both energy and momentum.)
Solution:
We must conserve both energy and relativistic momentum of the system of
fragments. With subscript 1 referring to the 0.868c particle and subscript 2 to
the 0.987c particle,
1 
1
1  0.868
2
 2.01 and  2 
1
1  0.987
2
 6.22 .
Conservation of energy gives E1  E2  Etotal
which is
or
This reduces to:
 1m 1c2   2m 2c2  m totalc2
2.01m 1  6.22m 2  3.34  1027 kg .
m 1  3.09m 2  1.66  1027 kg . (1)
Since the final momentum of the system must equal zero, p1  p2
 1m 1u1   2m 2u2
gives
 2.01 0.868c m 1   6.22 0.987c m 2
or
which becomes
m 1  3.52m 2 .
(2)
28
Solving (1) and (2) simultaneously, m 1  8.84  10 kg and
m 2  2.51 1028 kg .
4. A refrigerator has a coefficient of performance of 3.00. The ice tray
compartment is at –20.0°C, and the room temperature is 22.0°C. The
refrigerator can convert 30.0 g of water at 22.0°C to 30.0 g of ice at –
20.0°C each minute. What input power is required? Give your answer in
watts.
Q
CO P  3.00  c . Therefore,
W
The heat removed each minute is
Qc
W 
3.00 .


QC
  0.030 0 kg 4186 J kg C   22.0C    0.030 0 kg 3.33  105 J kg
t
  0.030 0 kg 2 090 J kg C   20.0C   1.40  104 J m in
or,
Qc
 233 J s .
t
Thus, the work done per sec
P 
233 J s
 77.8 W
3.00
.
5. The total volume of water in the oceans is approximately 1.40 × 109
km3. The density of sea water is 1 030 kg/m3, and the specific heat of the
water is 4 186 J/(kg · °C). Find the increase in mass of the oceans
produced by an increase in temperature of 10.0°C.
Solution:
m 
E

c2
m c T 
c2

V c T 
c2
m  6.71 108 kg
1030 kg m 1.40  10 10 m   4186 J kg C  10.0 C 

 3.00  10 m s
3
9
3
3
8
2
1. Time Dilation:
t  t p
Formulas:
L
2. Length Contraction:
Lp

3. Relativistic Momentum:
p
mu
1
u
2
 mu
E 2  E02
p 
c
c2
4. Relativistic Energy:
Rest Energy:
E0  mc2
E  K  E0 
Total Energy:
mc 2
1
u2
c2
mc2 (for electron) = 5.11 x 105eV
5. Relation between the total energy, momentum, and the rest energy
for a relativistic particle:
E 2  p 2 c 2  (mc 2 ) 2
E  pc,
for
E  mc2
6. The first Bohr’s radius (for electron in hydrogen atom in the ground
state):
a0 
2
2
 0.0529nm
mke
7. The ground energy level in hydrogen atom:
E0 
mk 2e 4
2 2
1 ke2

 13.6eV
2 a0
8. The energy levels, orbit radius, and kinetic energy of electron in an
atom:
rn 
E
En   Z 2 20
n
n 2 2
mke2
1 2 kZe2
K E  mv 
2
2r
 n a0
2
9. The reduced mass for an atom:

me mn
me  mn
where me is the mass of electron, mn is the mass of the nuclei
10.The Bragg condition:
2d sin   n
n  1,2,3.....
11. The Compton Effect:
C 
h
 
1 cos 
m ec
h
me c
12. Relativistic Doppler Effect:
v
v
1
- approching
c f- r - receding
c f
f '
f '
0
0
v
v
1
1
c
c
13. Useful Formulas for Speed, Energy, and Momentum:
1
E  hf 
hc

f 
c

u
pc

c
E
14. Photoelectric Effect:
 mv2 

eV0  
 hf  
 2 

 max
c 
hc

u ( ) 
8hc 25
hc
e kT
15. Plank’s Law:
1
R  T 4
16. Stefan-Boltzman Radiation Law:
17. Wien’s displacement law:
max T  cons tan t  2.898 103 m  K
18. Lorentz Transformation Equations:
x   ( x'vt)
x'   ( x  vt)
y  y' ,
y '  y,
z  z'
z'  z
v 

t    t ' 2 x' 
 c 
u v
u x'  x
u v
1  x2
c
u 1x  v
ux 
u 1x v
1 2
c
19. Specific Heat:
c
20. Latent Heat:
t'  t
Q
mT

Q

m
Wengine  Qnet
L
21. Thermal Efficiency: e 
Weng
Qh

22. Theoretical (Carnot) Efficiency:
Q  mcT
Q   Lm
Qh  Qc
Q
 1 c
Qh
Qh
e
Th  Tc
Tk
23. Coefficient
of
Performance 
QC
W
kB = 1.38 x 10-23J/K
me=9.11 x10-31kg=0.511MeV/c2
e = -1.6 x 10-19C
hc = 1240 eV∙nm = 1.98 x 10-25J∙m
ћ = 1.055x10-34J·s
h = 6.62 x 10-34J.s=4.136 x 10-15eV∙s
1u = 1.66x10-27kg
NA= 6.02 x 1023 mol-1
1J = 6.24x1018eV;
1eV = 1.602x10-19J
c = 3.00x108 m/s
mp = 1.673x10-27kg= 938.3 MeV/c2
k = 8.98x109 Nm2/C2 mn= 1.675x10-27kg=939.6 MeV/c2
σ = 5.672 x 10-8 W/m2K4