Hydrogen `(""_(1)H^(1))`, Deuterum `(""_(1)H^(2))`, singly ionised Hellium `(""_(2)He^(4))^(+)` and doubly ionised lithium `(""_(3)Li^(6))^(++)` all have one electron around the nucleus. Consider an electron tranition from `n=2` to `n=1`. If the wave lengths of emitted radiation are `lambda_(1),lambda_(2),lambda_(3)` and `lambda_(4)` respectively then approximately which one of the follwing is correct ?

Options

A. `4lamda_(1)=2lamda_(2)=2lamda_(3)=lamda_(4)`
B. `lamda_(1)=2lamda_(2)=3lamda_(3)=lamda_(4)`
C. `lamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`
D. `lamda_(1)=2lamda_(2)=3lamda_(3)=4lamda_(4)`

Answer

C

Explanation

For hydrogen, we get
`(1)/(lamda)=Rz^(2)((1)/(1^(2))-(1)/(2^(2)))implies (1)/(lamda_(1))=R(1)^(2)((3)/(4))`
`(1)/(lamda_(2))=R(1)^(2)((3)/(4))implies (1)/(lamda_(3))=R(2)^(2)((3)/(4))`
`(1)/(lamda_(4))=R(3)^(2)((3)/(4)) implies (1)/(lamda_(1))=(1)/(4lamda_(3))=(1)/(9lamda_(4))=(1)/(lamda-(2))`
`implies lamda_(1)=lamda_(2)=4lamda_(3)=9lamda_(4)`