Physics 1C
Week 7 Questions
Fall 1998
1. One of the mirrors of a Michelson interferometer is moved and 426 fringe pairs pass the centre
point of the viewing telescope. The instrument is illuminated with 632.8 nm wavelength light from
a He-Ne laser. How far was it moved?
2. A slit of width 10m is illuminated with light of  = 632.8 nm. Find the angular position of the
first diffraction minimum.
3. Light from a sodium lamp has two strong yellow components at 589.5923 nm and 588.9953 nm.
How far apart in the first-order spectrum will these two be on a screen 1.00 m from a grating
having 10,000 lines/cm?
4. 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
5. An alpha particle with a kinetic energy of 4.05 MeV is fired directly at a Ge32 nucleus. How close
does it get?
Physics 1C
1.
2.
Week 7 Solutions
Fall 1998
x
The path difference increases by  for every /2 moved by our
mirror and results in one fringe passage. Thus, x = N(/2) =
426/2 (632.8 nm) = 0.135 mm.
 =
N
N

2
Minima occur at asin = m, n = 1, 2, 3, …
a

At the first minimum m = 1
m 
 ,
a
a
(632.8  10 9 )

 = arc sin   = arc sin
= 3.628 o
(10  10 6 )
a
Solve for ,
sin =
3.
y1 = s1/a, y2 = s2/a, y1 - y2 = (589.5923 - 588.9953)(10-9)(1.00 m)/(10-6) = 0.597 mm. Using s = 1m and a = 10 -4
cm = 10-6 m.
4.
(a) E = 511 + 60 = 571 keV
(b) E  p c  mo c  p 
2
2
2
2
4
1
1 2
( E  mo2 c 4 ) 2
c
1
1
1
p  (5712  5112 ) 2  (254.8)  254.8 keV/c
c
c
(c)
E  mc 2 
mo c 2
1
m c 
v
 2   o 
c
 E 
2
2
(d)
m
mo
v2
1 2
c

v2
c2
2
 1
v 2 mo c 2

E
c2
1
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)
5.
At closest approach, initial KE = work done = PE =
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
x 
6
19
(4.05  10 )(1.6  10 )
KE = 4.05 MeV
KE = 0
at closest approach



 particle
x
Ge32
nucleus