Calculate the wavelength and energy of radiation emitted for the electronic transition from infinity `(oo)` to stationary state one of the hydrogen atom `("R = 1.09678 "xx 10^(7) m^(-1))`

Answer

A, B, C, D

Explanation

`(1)/(lambda) = R [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]`
` = 109678 cm^(-1)[(1)/(1^(2)) - (1)/(oo^(2))]`
`= 109678 cm^(-1)`
`lambda = (1)/(109678 cm^(-1)) = 9.118 xx 10^(-6) cm`
`E = (hc)/(lambda) = (6.62 xx 10^(-34) xx 3 xx 10^(8))/(9.118 xx 10^(-8)) = 2.178 xx 10^(18) J`