Calculate the wavelength and energy of radiation emitted for the electron transition from infinite `(oo)` to first stationary state of the hydrogen atom.
`R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s ` and `c = 2.979 xx 10^(8) m s^(-1)`

Answer

`lambda = 9.11 xx 10^(-8) m, E = 2.18 xx 10^(-18) J`

Explanation

According to Rydberg equation,
`1/lambda = 109678 (1/(n_1^2) - 1/(n_2^2))`
Here, `n_1 = 1, n_2 = oo`
`:. 1/lambda = 109678 (1/(1^2) - 1/(oo^2))`
`1/(lambda) = 109678`
or `" " lambda = 1/(109678) = 9.11 xx 10^(-6) cm`
or `" " = 9.11 xx 10^(-8) m`.
Now, `E = (6.63 xx 10^(-34) xx 3.0 xx 10^8)/(9.11 xx 10^(-8))`
`= 2.183 xx 10^(-18) J`.