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

Explanation

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,
`(1)/(lambda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))`
`"so, "(1)/(lambda)=R((1)/(2^(2))-(1)/(2^(2)))=(5R)/(36) and (1)/(lambda.)=R((1)/(3^(2))-(1)/(4^(2)))=(7R)/(144)`
`therefore" "lambda.=(144)/(7)xx(5lambda)/(36)=(20lambda)/(7)`