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 form the `4th` orbit to the `3dr` orbit, the corresponding wavelength of the photon will be

Options

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

Answer

C

Explanation

`(1)/(lamda)=R(1)^(2)((1)/(2^(2))-(1)/(3^(3)))`
`(1)/(lamda_(2))=R(1)^(2)((1)/(3^(2))-(1)/(4^(2)))`
`((1)/(lamda))/((1)/(lamda_(2)))=((1)/(4)-(1)/(9))/((1)/(9)-(1)/(16))=((5)/(4xx9))/((16-9)/(9xx16))`
`(lamda_(2))/(lamda)=(5)/(4xx9)xx(9xx16)/(7)=(20)/(7)`
`lamda_(2)=(20)/(7)lamda`