Q Current.doc

1
(a)
Estimate the energy of
the electrons that we
need to use in an
electron microscope to
resolve a separation of
0.27 nm.
(b)
In a scattering of 2 eV
protons from a crystal,
the fifth maximum of the
intensity is observed at
an angle of 300
. Estimate
the crystal’s planar
separation.

2
Solution
 
We know that each material particle
of momentum behaves as a group
of waves matter waves whose
wavelength and wave vector
are given as
= , and =
where
2
This is known as the de Broglie
p
k
h p
k
p
h








relation which connects the
momentum of a particle with the
wavelength and wave vector of
the wave corresponding to this
particle.

3
2
2
2
2 2 2
2 2 2
2
2
2 18 2
6
( )
Let be the kinetic energy
of the electron and be the
mass of the electron, then
1
2
1
2 ( )
( )
2 2 2
(1240 . )
2 0.511 (0.27 nm)
1537600 10
0.07 10 10
e
e
e
e
e e e
a
E
m
E m v
p
m X
m
p h hc
m m m c
eV nm
X MeVX
XeV X m
X eVX
 




  

 18 2
20.63
m
eV


4
2
2
2
2 2
2
( )
Let be the kinetic energy
of the electron and be the
mass of the proton, then
1
2
1
2 ( )
(1)
2 2
p
p
p
p
p p
b
E
m
E m v
p
m X
m
p h
m m 


    
/
/

5
2 2
2 2
2 2
2
2
Using Bragg's relation,
2
= sin ,
where d is the crystal
planar separation,
in eq.(1), we get
2 4 sin
...(2)
8 sin
p
p
d
n
h n
E X
m d
n h
d
E m
 



 
/
/
15
2
Using
5, 30 , 2 ,
4.136 10 . ,
938.3 ,
in eq.(2), we get
p
n E eV
h X eV s
MeV
m
c


  


/

6
30 2 2 2
2
6
30 16 2
6
20 2
10
9
25 17.11 10
2 8 938.3 10 0.25
427.75 10 9 10
3753.2 10
1.026 10
1.026 10
0.1026 10
0.1026
X X eV s Xc
d
eVX X X eVX
X X X m
X
X m
d X m
X m
nm








 



Q Current.doc

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