find process for radius of electron is equal to 0.529 ×n^2/z Å
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2
Answer:
Let us consider the n
th
bohr orbit,
r
n
=
4π
2
mZe
2
n
2
h
2
For hydrogen atom z=1, first orbit n=1
r
1
=
4πr
2
me
2
h
2
=0.592A
0
(i)For He
+
ion,Z=2,thirdorbit,n=3
r
3
(He
+
)=
4π
2
m×2×e
2
3
2
h
2
=
2
9
[
4π
2
me
2
h
2
]
=
2
9
×0.592
=2.380A
0
(ii) For Li
2+
ion,Z=3,secondorbitn=2
r
2
(Li
2+
)=
4π
2
m×3×e
2
2
2
h
2
=
3
4
[
4π
2
me
2
h
2
]
= 0.705A
0
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