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The time period of revolution of an elec...

The time period of revolution of an electron in its ground state orbit in a hydrogen atom is `1.6 xx 10^(-16)` s. The frequency of the revoltuion in ( ` s^(-1)`). of the electron in its second exited state is

A

`5.6 xx 10^(12)`

B

`7.8 xx 10^(16)`

C

`1.6 xx 10^(14)`

D

`2.3 xx 10^(14)`

Text Solution

Verified by Experts

The correct Answer is:
D
`T alpha (r)/(v) alpha (n^(2))/(z) .(n)/(z) `
So `Talpha (n^(3))/(z^(2)) , " " T_(3rd) = (3^(3))/(1^(3)) xx 1.6 xx 10^(-16)`
`27 xx1.6 xx 10^(-16)`
So `f = (1)/( 27 xx 1.6 xx 10^(-16)) = 2.3 xx 10^(14)`
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