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Find the quantum of the excited state of...

Find the quantum of the excited state of electrons in `He^(+)` ion which on transition to first excited state emit photons of wavelengths `108.5 nm`. `(R_(H)=1.09678xx10^(7) m^(-1))`

A

`6`

B

`5`

C

`4`

D

`2`

Text Solution

Verified by Experts

The correct Answer is:
B
`1/lambda=R_(H)Z^(2)[1/1^(2)-1/n_(2)^(2)]`
`1/(108.5xx10^(-9))=1.09678xx10^(7)xx4[1/2^(2)-1/n_(2)^(2)]`
This gives `n_(2)=5`
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