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 approximately
A. `1.5` eV
B. `0.85` eV
C. `3.4` eV
D. `1.9` eV