what are the frequency and wavelength of a photon emitted during a transition from n = 5 state to the n =2 state in the hydrogen atom?

Explanation

Since `n_(i) = 5` and `n_(f) = 2`, this transition gives rise to a special line in the visible region of the Bamer series.
`DeltaE = 2.18 xx 10^(-18)J [(1)/(5^(2))-(1)/(2^(2))] =- 4.58 xx 10^(-19)J`
It is an emission energy. The frequency of the photon (taking energy in terms of magnitude ) is given by
`v = (Delta E)/(h) = (4.58 xx 10^(-19)J)/(6.626 xx 10^(-34)Js) = 6.91 xx 10^(14)Hz`
`lambda = (c)/(v) = (3.0 xx 10^(8)ms^(-1))/(6.91 xx 10^(14)Hz) = 434 nm`