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The wavelengths of photons emitted by el...

The wavelengths of photons emitted by electron transition between two similar leveis in H and `He^(+)` are `lambda_(1)` and `lambda_(2)` respectively. Then :-

A

`lambda_(2)=lambda_(1)`

B

`lambda_(2)=2lambda_(1)`

C

`lambda_(2)=lambda_(1)//2`

D

`lambda_(2)=lambda_(1)//4`

Text Solution

Verified by Experts

The correct Answer is:
D
`DeltaE=(hc)/(lambda)`
`DeltaE=(hc)/(lambda_(1))` (for H atom)
`DeltaE=xxƶ^(2)=(hc)/(lambda_(2))` (for `He^(+)`atom)
`(hc)/(lambda_(1))xx4=(hc)/(lambda_(2))implies lambda_(2)=(lambda_(1))/(4)`
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