DERIVE an expression for frequency of electron in nth orbit of H like atom
Answers
Answer:
From Bohr's postulates, angular momentum of an electron in n
th
orbit is given by:
L
n
=
2π
nh
∴m
e
v
n
r
n
=
2π
nh
................(i)
From Coulomb's Law,
F=
4πε
o
r
2
Q
1
Q
2
F=
4πε
o
r
n
2
e
2
as atomic number of hydrogen is 1, nucleus has only one proton
This provides the necessary centripetal force for the electron to remain in orbit.
r
n
m
e
v
n
2
=
4πε
o
r
n
2
e
2
..............(ii)
Substiuting value of r
n
from (ii) in (i), we get:
m
e
v
n
4πε
o
m
e
v
n
2
e
2
=
2π
nh
v
n
=
2nε
o
h
e
2
........(iii)
Now, kinetic energy of electron is:
KE
n
=
2
1
m
e
v
n
2
=
2
1
m
e
4n
2
ε
o
2
h
2
e
4
KE
n
=
8n
2
ε
o
2
h
2
m
e
e
4
...........(iv)
Potential energy of electron is:
PE
n
=−
4πε
o
r
n
Q
1
Q
2
=−
4πε
o
nh
e
2
2πm
e
v
n
using (i)
Substituting v
n
from (iii), we get:
PE
n
=−
2ε
o
nh
e
2
m
e
×
2nε
o
h
e
2
=−
4n
2
ε
o
2
h
2
m
e
e
4
.............(v)
Total energy is hence, given by:
E
n
=KE
n
+PE
n
=−
8n
2
ε
o
2
h
2
m
e
e
4
It can be observed that E
n
=−KE
n
Given E
1
=−X eV
⟹KE
n
=X eV
Explanation:
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