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An electron jumps from the 4^(th) orbit...

An electron jumps from the `4^(th)` orbit to the `2^(nd)` orbit of hydrogen atom. Given the Rydberg's constant `R=10^(5) cm^(-1)` The frequency in Hz of the emitted radiation will be

A

`3/16 xx10^5`

B

`3/16 xx10^(15)`

C

`9/16 xx10^(15)`

D

`3/4 xx10^(15)`

Text Solution

Verified by Experts

The correct Answer is:
C
`1/lamda=R(1/2^2-1/4^2)=(3R)/16=16/3xx10^(-5) cm`
Frequecy `c/lamda=(3xx10^(10))/(16/3xx10^-5)=9/16xx10^(15) Hz`
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