If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength `lambda`. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be

Options

A. `(16)/(25)lambda`
B. ` (9)/(16)lambda`
C. `(20)/(7)lambda`
D. `(20)/(13)lambda`

Answer

C

Explanation

Excess energy appears as photon.
From Rhydberg's formula,
`" " (1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))=R((1)/(2^(2))-(1)/(3^(2)))=(5R)/(36)`
`" " (1)/(lambda')=R((1)/(3^(2))-(1)/(4^(2)))=(7R)/(144)`
`" " (1)/(lambda)//(1)/(lambda')=(5R)/(36)-:(7R)/(144)`
`rArr" " (lambda')/(lambda)=(5R)/(36)xx(144)/(7R)=(20)/(7)rArrlambda'=(20)/(7)lambda`