PHYSICS 2D WINTER 2010 FINAL EXAM
SOLUTIONS
1 (a) Let us assume enemy spaceship (reference frame S) is approaching
your ship (reference frame S) from left with velocity v = 0.4 c along x-axis.
Velocity of missile in S = ux = 0.7c along x-axis.
Use reverse velocity transformation to get speed ux in reference frame S.
ux 
u x  v
0.7c  0.4c
1.1


c  0.86c
2
1  ux v / c
1  (0.7)(0.4) 1.28
(b) In reference frame S, missile was fired at distance L = 8 x 109 m
and is approaching with speed of 0.86c
so time taken in Reference frame S = (8 x 109 )/ (0.86 x 3 x 108) s = 31 secs
= T, say.
(c) In reference frame S, because you are moving with velocity v relative to
him, when he fired, your distance by his measurement was only 1 
v2
L
c2
from him, because of length contraction . If he estimates time T for missile
to reach you, you will have in his frame moved a distance vT towards him,
so
1
T 
T (1 
v2
L  vT 
c2
u x
v
v2
)  1  2 L / u x
u x
c
v2
L
2
c 2  1  (0.4) .8.10 9 s  22.2s
u x  v
(1.1)(3)10 8
1
T 
Alternatively, we can use the Lorentz transformation :
T   t    (t  vx / c2 )
where t = time between firing and arrival in frame S = T = 31 s from (b)
and x = distance between firing and arrival in frame S = L= 8.109 m
1
so T  
1  (0.4)
2
[31  (0.4)(8.10 9 )]s  22.2s
2. We have KE of electron = hf -  = h(c/) - 
= (4.136. 10-15) eV.s x ( 3. 108)/(350. 10-9) s-1 - 2 eV
= 3.52 eV – 2 eV = 1.52 eV
so stopping voltage = 1.52 Volts.
3. We have phase velocity v p 
Now  

k
(
2 S

)1/2 / 1/2
2

2 S 1/2 k 1/2
so  (
) ( )
k
k

2
S
so   ( )1/2 k 3/2
so vg 

d
dk

3 S 1/2 1/2 3 2 S 3
( ) k 
 vp
2 
2 
2
4. (a) En=3 = -13.6(1/32)(22) eV for He (Z=2)
E n=2 = -13.6(1/22)(22) eV
so photon energy = hf = (En=3 - En=2 ) = 13.6 x 4 x (1/4 -1/9) eV = 7.56 eV
so f = 7.56 eV/h = 7.56 eV/(4.136. 10-15) eV.s =1.83. 1015 s-1
so  = c/f = 3.108/1.83.1015 m = 1.64. 10-7 m
(b) rn=2 = 4a0/Z = 2a0 =1.06.10-10 m
5. (a) From solutions of SE for H atom, for n = 3, l= 0,1,2
Ignoring electron spin,
l=0 ; m= 0 s-state ( singly degenerate)
l=1 ; m = -1, 0, +1 p-state (triply degenerate)
l= 2; m =-2, -1, 0, +1, +2 d-state (5 –fold degenerate)
Total degeneracy of (n=3) state = 9
(b) For 3d state, l =2 , so L  h l(l  1)  6h
max. value of m = +2, so max. value of Lz  2h
6. (a) For infinite potential 1D box,  n (x)  Asin(kn x) and must vanish for
x=0 and x=L, so kn  n / L
Thus,  n (x)  Asin(n x / L) where A 
2
L
  (x)
condition
2
follows from the normalization
L
dx  1
0
2
sin(n x / L)
L
h2 k 2 h2  2 2
( ) n
and En  n 
2m
2m L
so  n (x) 
Ground state follows by putting n=1.
(b) For n=1,
L
L
L
2
2 1
x    *1 (x)x 1 (x)dx  ( )  sin 2 ( x / L)xdx  ( )  x (1  cos(2 x / L))dx
L 0
L 02
0
1 1
L
L
 ( )[ x 2  x
sin(2 x / L)  ( )2 cos(2 x / L)]0L  L / 2
L 2
2
2
L
x
2
L
2
2
1
 ( )  dxx 2 sin 2 ( x / L)  ( )  dxx 2 (1  cos(2 x / L))
L 0
L 0
2
2 1
1
L
1
L
1
 ( )[ x 3  x 2 ( )sin(2 x / L)  .2 
cos(2 x / L)]0L  L2
L 6
2
2
2
2
3
L
p  (i2 / L)h  dx sin( x / L)
0
d

(sin( x / L))  (i2 / L)h( )  dx sin( x / L)cos( x / L)  0
dx
L 0
L
2
d2
2 2 h2
 2 h2
p 2  ( )(ih)2  dx sin( x / L) 2 (sin( x / L))  ( 3 )  dx sin 2 ( x / L)  2
L
dx
L
L
0
0
L
L
(c)
x 
x2  x
p 
p2  p
xp  (

12
2
2
 L2 / 3  (L / 2)2  L / 12

h
L
)h  0.9h