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In the Davisson-
Germer experiment
using a Ni crystal, a
second-order beam
is observed at an
angle of 55◦
. For
what accelerating
voltage does this
occur?
HELPING TOOLS
No.1
2
No.2
No.3
3
No.4
4
Solution
We know that maxima for a diffraction
grating occur at angles such that the
path difference between adjacent rays
sin is equal to a whole number of
wavelengths:
sin 1,2,3,... (1)
where is
d
d n n
n


 
= = − − − −
the order number of the
maximum.
From independent data, it is known
that the spacing between the rows of
atoms in a nickel crystal is
0.215 .
d nm
=
5
Using
2, 55 , 0.215
in eq.(1), we get
0.215n sin 55 2
0.215n sin 55
2
0.215 0.82
2
0.08815
n d nm
mX
mX
nmX
nm







= = =
=
 =
 =
=
8
15
9
2
,
2.998 10
4.136 10 .
0.0881 10
140.746 10
14074.6
Also
E hf h h
p p
c c
m
X
X eV s s
p X
X m c
eV
p X
c
eV
p
c
 
−
−
= = =  =
 =
 =
 =
6
2
2
2
2
2
2
,
1
K &
2 2
(14074.6 )
2
198094365.16( )
2
198094365.16( )
2 511000
193.83
Now
p
mv p mv K
m
eV
c
K
m
eV
K
mc
eV
K
X eV
K eV
= =  =
 =
 =
 =
 =
This desired KE, i.e. 193.83
must be achieved & for achieving
this one, the electrons must be accelerated
through a PD of 193.83 as
electric PE & potential are related by
eV
V V
U q V
 = +
 = 

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Davison German Exp.pdf

  • 1. 1 In the Davisson- Germer experiment using a Ni crystal, a second-order beam is observed at an angle of 55◦ . For what accelerating voltage does this occur? HELPING TOOLS No.1
  • 4. 4 Solution We know that maxima for a diffraction grating occur at angles such that the path difference between adjacent rays sin is equal to a whole number of wavelengths: sin 1,2,3,... (1) where is d d n n n     = = − − − − the order number of the maximum. From independent data, it is known that the spacing between the rows of atoms in a nickel crystal is 0.215 . d nm =
  • 5. 5 Using 2, 55 , 0.215 in eq.(1), we get 0.215n sin 55 2 0.215n sin 55 2 0.215 0.82 2 0.08815 n d nm mX mX nmX nm        = = = =  =  = = 8 15 9 2 , 2.998 10 4.136 10 . 0.0881 10 140.746 10 14074.6 Also E hf h h p p c c m X X eV s s p X X m c eV p X c eV p c   − − = = =  =  =  =  =
  • 6. 6 2 2 2 2 2 2 , 1 K & 2 2 (14074.6 ) 2 198094365.16( ) 2 198094365.16( ) 2 511000 193.83 Now p mv p mv K m eV c K m eV K mc eV K X eV K eV = =  =  =  =  =  = This desired KE, i.e. 193.83 must be achieved & for achieving this one, the electrons must be accelerated through a PD of 193.83 as electric PE & potential are related by eV V V U q V  = +  = 