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

Answer

B

Explanation

`(1)/(lambda)=Rz^(2)((1)/(1^(2))-(1)/(2^(2)))`
`(1)/(lambda_(1))=R(1)^(2)(3//4) (1)/(lambda_(2))=R(1)^(2)(3//4)`
`(1)/(lambda_(3))R2^(2)(3//4)(1)/(lambda_(4))=R3^(2)(3//4)`
`(1)/(lambda_(1))=(1)/(4lambda_(3))=(1)/(9lambda_(4))=(1)/(lambda_(2))`
So option `(3)` is correct