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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








 



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Q Current.doc

  • 1. 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. 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. 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. 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. 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. 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            