Physics 1C
Week 8 Questions
Fall 1998
1. An electron of rest energy 511 keV is accelerated through 60kV in a HDTV set. Find its:
(a) total energy E
(b) momentum P
(c) speed v
(d) mass m, measured by an observer at rest
2. An alpha particle with a kinetic energy of 4.05 MeV is fired directly at a Ge32 nucleus. How close
does it get?
3. Find the wavelength of an electron, a proton and a photon each of kinetic energy 200 keV.
4. (a) Compute the amount of energy liberated in the fusion reaction:
2
H + 2H  3He + 1n + energy.
Here, the neutron flies off with about 3 times the KE of the helium.
(b) Another important fusion reaction involving deuterium and tritium is:
2
H + 3H  4He + 1n + energy
where the neutron flies off with about 4 times the KE of the helium. Determine the total
amount if energy liberated.
5. Recall that a mass spectrometer separates singly charged ions travelling with velocities v
perpendicular to a magnetic field B into circular paths of radii r = mv/eB, where e is the electronic
charge and m is the mass of an ion. Three such (molecule X and elements Y and Z) are accelerated
through 1 kV and then enter a field of 0.5 T. The radii are 60.43, 80.98 and 127.87 mm
respectively. From the Periodic Table identify X, Y and Z.
6. A tiny droplet of oil carries an extra charge of e and is held motionless between parallel plates 2.00
cm apart, across which there is a potential difference of 60.0 kV. What is the mass of the droplet?
7. A beam of X-rays having a wavelength of 0.090 nm impinges on an unknown crystal. The
strongest reflection maximum is observed to occur at an angle of 25o. What is the spacing between
the atomic planes doing the scattering?
Physics 1C
Week 8 Solutions
Fall 1998
1. (a) E = 511 + 60 = 571 keV
1
1
(b) E 2  p 2 c 2  mo2 c 4  p  ( E 2  mo2 c 4 ) 2
c
1
1
1
p  (5712  5112 ) 2  (254.8)  254.8 keV/c
c
c
2
mo c
v2 m c2
(c) E  mc 2 
 1 2  o
E
c
v2
1 2
c
m c 
v
 2   o 
c
 E 
2
2
(d) m 
mo
v2
1 2
c
2

1
2. At closest approach, initial KE = work done = PE =
x 
1
  m c2 2  2
  511  2  2
o
  c  1  
hence v  1  
  c  0.446 c
  E  
  571  
511
511
511


 571 keV/c2
2
0
.
895
1  0.199
1  (0.446)
kQ( 2e)
,
x
x
kQ( 2e)
KE
(9  10 9 )(32)(1.6  10 19 )2(1.6  10 19 )
 2.28  10 14  22.8 fm
6
19
(4.05  10 )(1.6  10 )
KE = 0
at closest approach
KE = 4.05 MeV



 particle
Ge32
nucleus
x
3. (a) The e- rest mass (energy) is 512 keV. KE is not  moc2, and thus it is relativistic.
E 2  p 2c 2  mo2c 4  p 2c 2  E 2  mo2c 4  p 

h

p
hc
E 2  mo2 c 4

4.14  10 keV s3  10
712  512 keV 
18
2
2
Page 1
2
1
8
2
m/s
1
E 2  mo2c 4
c
  1.242  10
494.77
9
 2.51 pm
Physics 1C
Week 8 Solutions
Fall 1998
(b) p+ rest energy is 938 MeV. KE is  moc2, and thus it is not relativistic.

h

p
h
2mk

4.14  10 18 keV s
 

3 keV 
2 938  10 c 2 200 keV 

 


OR
1

(4.14  10 18 )3  10 8
2
1.876  10 200
6
1
 6.41  10 14 m  64 fm
2
6.626  10 34 J s


31
5
19 J  
2 9.1  10 kg 2  10 eV 1.6  10 eV 






1
 2.75 pm
2
(c) photon (h) is completely relativistic, c = .


18
8 m
hc 4.14  10 keV s 3  10 s
E  h 
 

 6.21 pm

E
200 keV
hc
4. (a) 2(1876.12 MeV) - 2809.41 MeV - 939.566 MeV = 3.26 MeV.
(b) 1876.12 MeV +2809.43 MeV - 3728.40 MeV - 939.566 MeV = 17.58 MeV.
mv 2
p 2 e 2 r 2B 2
5.
 evB  p  erB and K 

 eV ,
r
2m
2m
A


eB 2 r 2
1.6  10 19 0.5 r 2

 12048r 2
 27
2uV
2 1.66  10 1000

2

Y
Ax (mol. weight)  12048(0.06043) 2
= 44 (probably CO2)
Ay  12048(0.08098) 2  79 (Se)
Z
Az  12048(0.12787) 2  197 (Au)
X
6. mg = NeE; m = eE/g = e(V/d)/g = 4.90  10 14 kg.
7. d = 1(0.090 nm)/(2 sin25o) = 0.11 nm.
Page 2
er 2 B 2
the n m  Au 
2V