Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level. In which part of the electromagnetic spectrum does this line lie ?

Calculate the wavelength of the radiation emitted when an electron in

1.	Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level. In which part of the electromagnetic spectrum does this line lie ?
Answer» Solution :For hydrogen atom, `E_(N) - (21.8 xx 10^(-19))/(n^(2)) J "atom"^(-1)` Energy emitted when the electron jumps from `n = 4 " to " n = 2` will be given by `Delta E = E_(4) - E_(2) = 21.8 xx 10^(-19) ((1)/(2^(2)) - (1)/(4^(2))) = 21.8 xx 10^(-19) xx (3)/(16) = 4.0875 xx 10^(-19) J` The wavelength corresponding to this energy can be calculated using the EXPRESSION, `E = hv - h (c)/(lamda) ( :' c = V lamda)` so that `lamda = (hc)/(E) = ((6.626 xx 10^(-34) Js) (3 xx 10^(8) m s^(-1)))/((4.0875 xx 10^(-19) J)) = 4.863 xx 10^(-7) m = 4863 Å` (or 486.3 nm) It lies in the visible region