Calculate the energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 ina hydrogen atom.

Explanation

`barv = 109677 (1/(n_1^2) - 1/(n_2^2)) cm^(-1)`
For `n_1 = 2, n_2 = 3`
`barv = 109677 (1/(2^2) - 1/(3^2))`
`= 109677 xx 5/36 = 15232.9 cm^(-1)`
`Delta E = hv`
`= (hc)/(lambda) = hc barv`
`= (6.626 xx 10^(-34) Js)(3.0 xx 10^(10) cm s^(-1)) xx (15232.9 cm^(-1))`
`= 3.028 xx 10^(-19) J`
`v = c barv`
`= (3.0 xx 10^(10) cm s^(-1)) xx 15239.9 cm^(-1)`
`= 4.57 xx 10^(14) s^(-1)`.