The transition from the state `n = 4 to n = 3 ` in a hydrogen - like atom results in ultraviolet radiation, infrared radiation will be obtained in the transition

Options

A. `2 to1`
B. `3 to1`
C. `4 to2`
D. `5 to`

Answer

B

Explanation

We know that `(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))` for hydrogen atom for `e` transition from `n_(2) to n_(1)`
For `4 to 3, E prop (1)/(lambda) = R ((1)/(9) - (1)/(16)) = (7)/(9 xx 16) R`
For `2 to 1, 3 to 2`, we get more value of `1// lambda `i.e. , more energy than `4 to 3`
`IR` radiation has less energy than `UV` radiation.