When the electron in the hydrogen atom jumps from `2nd` orbit to 1st orbit, the wavelength of radiation emitted is `lambda`. When the electrons jumps from `3rd` orbit to `1st` orbit, the wavelength of emitted radiation would be

Options

A. `(27)/(32) lambda`
B. `(32)/(27) lambda`
C. `(2)/(3) lambda`
D. `(3)/(2) lambda`

Answer

A

Explanation

`(1)/(lambda) = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
First condition `(1)/(lambda) = R [(1)/(1^(2)) - (1)/(2^(2))] rArr R = (4)/(3 lambda)`
Second condition `(1)/(lambda') = R [(1)/(1^(@)) - (1)/(3^(2))]`
`rArr lambda' = (9)/(9R) rArr lambda' = (9)/(8 xx (4)/(3 lambda)) = (27 lambda)/(32)`