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The frequency of revolution of electron ...

The frequency of revolution of electron in first excited state in Hydrogen is `7.8xx10^(14)s^(-1)`. The frequency of revolution of electron is ground state in `H^+` will be close to (in `s^(-1)` ):

A

`6.2xx10^(15)`

B

`3.1xx10^(15)`

C

`1.2xx10^(16)`

D

`2.5xx10^(16)`

Text Solution

Verified by Experts

The correct Answer is:
D
`because fprop(z^(2))/(n^(3))`
`f_(H)=R(1^(2))/(2^(3))=(R)/(8)` . . . (i)
and `f_(He^(+))=R(2^(2))/(1^(3))=4R` . . . (ii)
`therefore f_(He^(+))=32f_(H)=32xx7.8xx10^(14)=249.6xx10^(14), `
`~~2.5xx10^(16)`.
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