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Hydrogen `(H)`, deuterium `(D)`, singly ionized helium `(He^(+))` and doubly ionized lithium `(Li)` all have one electron around the nucleus. Consider `n = 2` to `n = 1` transition. The wavelength of emitted radiations are `lambda_(1), lambda_(2), lambda_(3)` and `lambda_(4)` respectively. then approximately

A

`lambda_(1)=lambda_(2)=4 lambda_(3)=9 lambda_(4)`

B

`4lambda_(1)=2lambda_(2)=2lambda_(3)=lambda_(4)`

C

`lambda_(1)=2lambda_(2)=2sqrt(2)lambda_(3)=3sqrt(2)lambda_(4)`

D

`lambda_(1)=lambda_(2)=2lambda_(3)=3sqrt(2)lambda_(4)`

Text Solution

Verified by Experts

The correct Answer is:
A

`(1)/(lamda) prop Z^(2) implies lamda prop (1)/(Z^(2))`
`lamda_(1) : lamda_(2) : lamda_(3) :lamda_(4)=(1)/(1^(2)) :(1)/(1^(2)) :(1)/(2^(2)) : (1)/(3^(2))=1: 1:(1)/(4):(1)/(9)`
`lamda_(1)=lamda_(2)=4 lamda_(3)=9lamda_(4)`
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