King Fahd University of Petroleum and Minerals
Department of Physics
PHYS 212-MODERN PHYSICS
Term 071
Second Major Exam Solution
Sunday 02 December 2007
05:30-06:20 pm
Name:
ID No.:
1
1. Calculate the de Broglie wavelength for a proton moving at 106 m/s.
We have

h
h
6.626  10 34 J .s


 3.96  10 13 m  396 fm
p mv 1.673  10  27 kg  10 6 m.s 1
2. A proton has a kinetic energy of 1.0 MeV. If its momentum is measured with an
uncertainty of 5%, what is the minimum uncertainty in its position?
We have
p2
K
 p  2 Km
2m
We also have
px 


 x 
2
2p
Moreover,
p
 0.05  p  0.05  p  0.05  2 Km
p
Therefore,
x 

2  0.05  2Km

c
0.1  2Kmc 2

197.3eV .nm
0.1 2  10 6  938.3  10 6 eV 2
 46 fm
3. A free electron has a wavefunction
 x   A sin 5  1010 x 
where x is measured in meters. Find (a) the electron’s de Broglie wavelength, (b) the
electron’s momentum, and (c) the electron’s energy in eV.
We have
(a) k  5  1010 m 1 
(b) p 
h

2

 
2
2  3.14

   0.1256nm
k
5  1010 m 1
 k  1.055  10 34 J .s  5  1010 m 1  5.275  10  24 kgm / s
2
hc 
1240.8 eV 2 nm 2
p2


 95.5eV
2m 2  2  mc 2 2  0.12562 nm 2  0.511  10 6 eV
2
(c) K 
2
4. A proton is confined to moving in a one-dimensional box of width 0.200 nm. (a)
Find the lowest possible energy of the proton. (b) What is the lowest possible energy
of an electron confined to the same box? (c) How do you account for the large
difference in your results for (a) and (b)?
We have
 2 2
E1 
2mL2

 2 c 2
2mc 2 L2
(a) For a proton
E1 
 2 197.32 eV 2 nm 2
2  938.3  10 eV  0.2  nm
2
6
2
 5.11  10 3 eV
(b) For an electron
E1 
 2 197.32 eV 2 nm 2
2  0.511  10 6 eV  0.2  nm 2
2
 9.38eV
(c) According to Heisenberg’s uncertainty principle, for the same x , and therefore
the same  p , the particle with the smaller mass will have associated with it a larger
uncertainty on its speed. It will be moving faster and therefore will have more
kinetic energy than the particle with a larger mass.
5. Find the points of maximum and minimum probability density for the nth state of
a particle in a one-dimensional box. Check your results for the n = 2 state.
The probability density for the nth state is
 x  
2
2
 nx 
 sin 2 

L
 L 
It is maximum when
nx  3 5
L 3L 5 L
 , ,
,...,  x 
, , ,... ,
L
2 2 2
2n 2n 2n
It is minimum when
nx
L 2L
 0,  ,2 ,  x  0, ,
,...
L
n n
3
Of course 0  x  L in all cases.
For the n = 2 state we get
Maxima at x 
L 3L
,
4 4
L
Minima at x  0, , L
2
6. Write down the time-dependent Schrodinger’s equation for a particle of mass m
moving in a region where the potential energy is U(x).

2 2

x, t   U x  x, t   i x, t 
2
2m x
t
7. If x, t  is the wavefunction describing a particle at position x at time t. Write
down an expression for the probability of finding the particle in the interval from
x  a to x  b at time t.
2
b
P     x, t  dx
a
8. What is a stationary state?
It is a state for which the probability density is independent of time.
9. When do we say that an observable has sharp values?
When the state function of the system is an eigenfunction of the operator associated
with that observable.
10. Discuss the minimum energy of a quantum system using Heisenberg’s
uncertainty principle.
According to Heisenberg’s uncertainty principle, the minimum energy of a quantum
state cannot be zero because the particle is constantly moving since there is a
minimum uncertainty on its momentum.
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