A hydrogen atom emits ultraviolet of wavelength `102.5 nm` what are the quantum number of the state involved in the transition?

Answer

A, C, D

Explanation

As the light emitted lies in ultraviolet range, the line in lyman series
`1/lambda = R((1)/(n_1^2)-(1)/(n_2^2))`
implies `(1)/(102.5 xx 10^-9) = 1.1 xx 10^7((1)/(1^2)-(1)/(n_2^2))`
implies ` 10^9/102.5 = 1.1 xx 10^7((1)-(1)/(n_2^2))`
implies `10^2/102.5 = 1.1(1-(1)/(n_2^2))`
implies `1-(1)/(n_2^2) = (100)/(1025 xx 11 )`
implies `(1)/(n_2^2)=1-(100)/(102.5 xx 1.1 )`
implies `(1)/(n_2^2)=1-(100)/(102.5 xx 1.1)`
implies `n_2 = sqrt((112.75/12.75)) =3`
The transition is made from 1 to 3.