The document compares the frequency of revolution of an electron to the frequency of photons emitted during transitions from energy level n to n-1 for different values of n. For n=10, both frequencies are approximately 6.6x10^10 Hz. For higher n values like n=100, 1000, and 10,000, the radiation frequency remains the same at 6.6x10^10 Hz, while the revolution frequency decreases with increasing n.

1. 1
Compare the frequency
of revolution of an
electron with the
frequency of the
photons emitted in
transitions from n to
n − 1 for
(a) n = 10;
(b) n = 100;
(c) n = 1000;
(d) n = 10,000.

2. 2
HELPING TOOLS
No.1
Frequency of Revolution
No.2
Radiation Frequency
No.3

3. 3
No.4
No.5
No.6

4. 4
No.7
No.8
No.9
No.10
2 19 2
12
0
12 2
38
2
26
(1.602 10 )
4 4 3.14 8.85 10
2.57 10
111.156 10
2.57
111.156 10
e X C
F
X X X
m
X C m
X F
C m
X F

−
−
=
=
=

5. 5
2
26
26
26 18
8
9
0.023121 10
0.023121 10
0.023121 10 6.24 10
0.14427504 10 .
1.4427504 10 .
1.4427504 .
C m
X
C
Volt
X CmVolt
X X X eVXm
X eV m
X eV m
eV nm
−
−
−
−
−
=
=
=
=
=
=
Solution
4
3 2 3 3
0
(a) We know that frequency
of revolution is given by
---(1)
32
n
me
f
n
 
=
2 2 2
2 3
0 0
2
10 2
15
15 15
15 12
Re-writing Eq.(1) as
1 1
4 4 ( ) 2
511000
(1.440 . )
(197 . )
1
1000 4.136 10 .
1059609.6 10 1059609.6 10
4136 38809 160514024
0.0066 10 6.6 10
n
e e mc
f X X X X
c n
eV
f eV nm X
eV nm
X
X X eV s
X X
Hz
X s
X Hz X Hz
  
−
=
 =
= =
= =

6. 6
4
3 2 3 2 2
0
4
3 2 3 2 2
0
4
3 2 3
0
2 2
2
2
0
We know that radiation
frequency is given by:
1 1
( - )
64 ( 1)
2 1
= ---(2)
64 ( 1)
Using 10 in eq.(2)
19
64 8100
19
( )
4 4 ( ) 8100
me
f
n n
me n
n n
n
me
f X
e mc
X X
c
 
 
 
 
=
−
−
−
=
=
=
2
15
2
15
15
15 12
(1.44 . )
511000 19
4.136 10 . 8100
4 (197 . )
2
2.0736 511000 19
2600327188.8 10 .
20132582.4 10
2600327188.8
0.0077 10 7.7 10
MeV fm
eV
X X
X eV s
MeV fm X
X eVX
X eV s
X
Hz
X Hz X Hz


−
−
=
=
=
= =

7. 7
4
3 2 3 3
0
12
10
3
12
100 3
12
15 6
9
(b) From
32
6.6 10
10
6.6 10
100
1000
6.6 10
1000000
6.6 10 10
6.6 10
n
me
f
n
f X Hz
f X HzX
X X Hz
X X Hz
X Hz
 
−
=
=
 =
=
=
=
4
3 2 3 3
0
12
10
3
12
1000 3
12
15 9
6
(c) From
32
6.6 10
10
6.6 10
1000
1000
6.6 10
1000000000
6.6 10 10
6.6 10
n
me
f
n
f X Hz
f X HzX
X X Hz
X X Hz
X Hz
 
−
=
=
 =
=
=
=

8. 8
4
3 2 3 3
0
12
10
3
12
10000 3
12
15 12
3
(d) From
32
6.6 10
10
6.6 10
10000
1000
6.6 10
1000000000000
6.6 10 10
6.6 10
n
me
f
n
f X Hz
f X HzX
X X Hz
X X Hz
X Hz
 
−
=
=
 =
=
=
=