The wavelength of the radiations emitted when in a hydrogen atom electron falls from infinity to stationary state is ` : (R_H = 1. 097 xx10^7 m^(-1))` .

Options

A. `406 nm`
B. `192 nm`
C. `91 nm`
D. `9.1 xx 10^-8 nm`

Answer

C

Explanation

( c) `(1)/(lamda) = R[(1)/(n_1^2) - (1)/(n_2^2)]`
`(1)/(lamda) = 1.097 xx 10^7 m^-1 [(1)/(1^2) - (1)/(oo^2)]`
`:. lamda = 91 xx 10^-9 m`
We know `10^-9 = 1nm` So `lamda = 91 nm`.