Physics 1C
Week 10 Questions
Fall 1998
1. 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.
2. X-ray photons of E = 62 keV are Compton scattered at  = 120o from free electrons. Find the
scattered wavelength ’.
3. Derive the binding energy in eV of a hydrogen atom in its ground state where an electron of charge
-e is in circular orbit of radius 53 pm around a stationary proton of charge +e. This is not the full
Bohr derivation. It is much shorter and much easier.
k = 9  109
e = 1.6  10-19 C
4. Determine the wavelengths of the 3 longest wavelength lines H, H, H in the Balmer series for
H. What colors are these?
Physics 1C
1.
Week 10 Solutions (p.1)
mv 2
p 2 e 2 r 2B 2
 evB  p  erB and K 

 eV ,
r
2m
2m
A


er 2 B 2
2V

2

Y
Z
Az  12048(0.12787) 2  197 (Au)
2. The wavelength of 62 keV photons is obtained from E =

the n m  Au 
eB 2 r 2
1.6  10 19 0.5 r 2

 12048r 2
2uV
2 1.66  10  27 1000
Ax (mol. weight)  12048(0.06043) 2
= 44 (probably CO2)
Ay  12048(0.08098) 2  79 (Se)
X
Fall 1998
15
hc 4.136  10 ev s (3  10 ms )

 20.0129 pm
E
62  10 3 eV
8
hc

-1
hv
mo
In a Compton Effect scattering:
h
'   
(1  cos  )
mo c

6.626  10 34
 '  20.0129  10 
(1  cos 120  )
31
8
9.1  10 (3  10 )
 '  20.0129  2.4271 pm [1  (0.5)]
12
120o
hv'
'
 '  20.0129  3.6407
 '  23.6536 pm
3. The centripetal force is supplied by the electrostatic force
 mv 2
 ke2
ke2
Fc 
 FE  2  2 K  mv 2 
a
a
a
i.e. the kinetic energy is K 
ke2
2a
eFc
+e
a = 53 pm
The potential energy is PE  qV  (e)
ke
ke

a
a
2
Physics 1C
Week 10 Solutions (p.2)
ke2 ke2
The binding energy is the total energy U = KE +PE =

2a
a
2
2
9
- ke
- ke  1 
 ke  9  10 (1.6  10 19 )
U
(Joules) 

 13.6 eV
  eV 
2a
2a  e 
2a
2(5.3  10 11 )
4. The H levels are given by En =
 13.6
,
n2
n  1,2
Balmer series are transitions to n = 2
 H 
hc (4.136  10 15 )(3  10 8 ) 1.2408  10 6


 656.5 nm (red)
E
3.4  1.51
1.89
H 
1.2408  10 6
 486.6 nm (green)
3.4  0.85
 H
1.2405  10 6

 433.8 nm (blue)
3.4  0.54
-0.54
-0.85
-1.51
H
-3.4
-13.6
H
H
Fall 1998