Derive formula for energy of nth orbit at an atom (4m)
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Explanation:
ANSWER
We know,
mvr=
2n
nh
(i) [Bhor's theorem]
Also,
r
mv
2
=
r
2
kZe
2
- (ii)
We know in circular motion about electrostatic field
PE=−2KE
TE=PE+KE=−2KE+KE
TE=−KE
KE=
2
mv
2
From (i) and (ii), we have
r=
2nmv
nh
So, mv
2
=
r
kze
2
=
2nmv
nh
kze
2
⇒mv
2
=
nh
2nmvkze
2
⇒v=
nh
2nkze
2
So,
2
mv
2
=
2
m
[
nh
2nkze
2
]
2
2
mv
2
=
n
2
h
2
2n
2
k
2
z
2
e
2
m
2
mv
2
=[
h
2
2n
2
k
2
e
4
m
](
n
2
z
2
)
So, TE=−KE=
2
−mv
2
=[
h
2
−2n
2
k
2
e
4
m
](
n
2
z
2
)
We know,
For Hα line,
n
1
=2 and n
2
=3
λ
1
=R(z
2
)[
n
1
2
1
−
n
2
2
1
]
λ
1
=(1.1×10
7
)(1)[
4
1
−
9
1
]
λ
1
=0.513×10
7
=1.53×10
8
m
−1
λ=
153×10
8
1
=65.35 A
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