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

The energy emitted during transition from `n = 5` to `n = 2` state,
`Delta E = 2.18 xx 10^(-18) J (1/(n_i^2) - 1/(n_f^2))`
`= 2.18 xx 10^(18) (1/(5^2) - 1/(2^2))`
`= 2.18 xx 10^(-18) ((4 - 25)/(25 xx 4))`
`= -4.578 xx 10^(-19) J`
The negative sign indicates that the energy is emitted. For calculating wavelength and frequency, the minus sign may be omitted.
The frequency of emitted photon,
`c = (Delta E)/h = (4.578 xx 10^(-19) J)/(6.626 xx 10^(-34) J s)`
`= 6.91 xx 10^(14) s^(-1)`
`lambda = c/v = (3.0 xx 10^8 ms^(-1))/(6.91 xx 10^(14) s^(-1)) = 4.34 xx 10^(-7) m` or 434 nm