The wavelength of radiation emitted is `lambda_(0)` when an electron jumps from the third to the second orbit of hydrogen atom. For the electron jump from the fourth to the second orbit of hydrogen atom, the wavelength of radiation emitted will be

Options

A. `(25)/(16)lamda_(0)`
B. `(27)/(20)lamda_(0)`
C. `(20)/(27)lamda_(0)`
D. `(16)/(25)lamda_(0)`

Answer

C

Explanation

From the relation,
`(1)/(lamda_(0))=R((1)/(2^(2))-(1)/(3^(2)))=(5R)/(36)` . . . (i)
and also `(1)/(lamda)=R((1)/(2^(2))-(1)/(4^(2)))=(3R)/(16)` . . (ii)
From Eqs. (i) and (ii), we get
`(lamda)/(lamda_(0))=(5R//36)/(3R//16)=(20)/(27)implies lamda=(20)/(27) lamda_(0)`