YouTube Lecture # 23

1. 1
A certain crystal is cut so that
the rows of atoms on its
surface are separated by a
distance of 0.352 nm. A beam
of electrons is accelerated
through a potential difference
of 175 V and is incident
normally on the surface. If all
possible diffraction orders
could be observed, at what
angles (relative to the incident
beam) would the diffracted
beams be found?
Helping Tools
No.1
No.2

2. 2
No.3
No.4

3. 3
OR
No.5
OR

4. 4
Solution
2
2
2 2 2 2
2 2
2 2
2 2 2
4
We can have
1
K &
2 2
2 2
2 0.511 175
2 511000 175
178850000( )
13373.48
13373.48
1.337348 10
p
mv p mv K
m
p mK p c mc K
p c X MeVX eV
p c X eVX eV
p c eV
pc eV
eV
p
c
eV
p X
c
= =  =
 =  =
 =
 =
 =
 =
 =
 =

5. 5
15
4
19
19 8
11 9
,
4.136 10 .
1.337348 10
3.09 10 .
3.09 10 2.998 10 .
9.26 10 0.0926 10
0.0926
Also
E hf h
p
c c
h X eV s
eV
p X
c
X c s
m
X X X s
s
X m X m
nm


−
−
−
− −
= = =
 = =
=
=
= =
=

6. 6
sin
:
sin 1,2,3,.......(1)
where n i
We know that maxima for a
diffraction grating occur at
angles such that the path
difference between adjacent
rays d is equal to a
whole number of wavelengths
d n n


 
= =
1
s the order number of
the maximum.
sin
sin ................(2)
n
d
n
d



 −
 =
 =
1
1
1
2
1
3
Next, we solve eq.(2) for 1,2&3
For 1
For
sin ................(3)
2
sin ........
2
For 3
........(4)
3
sin ................(5)
n
d
d
d
n
n
n






−
−
−
=
=
=
=
=
=
=

7. 7
1
1
1
1
2
1
1
3
1
,
0.0926
sin
0.352
sin 0.263 15.248
2(0.0926)
sin
0.352
sin 0.526 31.736
3(0.0926)
sin
0.352
sin 0.789 52.092
Now
nm
nm
nm
nm
nm
nm



−
− 
−
− 
−
− 
=
= =
=
= =
=
= =