1.

Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximatelyA. `1.9 eV`B. `2.3 eV`C. `3.4 eV`D. `4.5 eV`

Answer» Correct Answer - A
`13.6 ((1)/(2^(2)) - (1)/(3^(2))) eV = 1.9 eV`


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