An electron jumps from the 3rd orbit to ...

An electron jumps from the 3rd orbit to the ground orbit in the hydrogen atom. If `R_H= 10^7` /m then the frequency of the radiation emitted in the transition is

`8/3xx10^15` Hz

`4/3xx10^15` Hz

`9/8xx10^15` Hz

`7/16 xx10^15` Hz

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The correct Answer is:

`1/lamda = R_H [(1)/(1^2) - (1)/(3^2) ] = 8/9 R_H`
`therefore v = v/lamda = 3xx 10^8xx 8/9xx 10^7`
`=8/3xx10^15 Hz`

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