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The number of revolutions done by an ele...

The number of revolutions done by an electron in one second in the first orbit of a hydrogen atom is of the ordr of 3.

A

`(u_(n))/(2pir_(n))`

B

`-(2E)/(nh)`

C

`-(2E_(1))/(n^(3)h)`

D

Any of these

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`u_(n)=(2pie^(2))/(nh)" "r_(n)=(n^(2)h^(2))/(4pi^(2)me^(2))`
`E_(n)=(2pi^(2)me^(4))/(n^(2)h^(2))`
Orbital frequency `=(u_(n))/(2pir_(n))`
`=(2pie^(2)xx4pi^(2)me^(2))/(nhxx2pixxn^(2)h^(2))=(4pi^(2)me^(4))/(n^(3)h^(3))`
`=(-2E_(n))/(nh)` (As `E_(n)=-(2pi^(2)me^(4))/(n^(2)h^(2)))`
`=(2E_(1))/(nhxxn^(2))=-(2E_(1))/(n^(3)h)`
Hence d is the correct answer.
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