Energy E of a hydrogen atom with principal quantum number n is given by $E=\frac{-13.6} {n^{6}}eV$. n The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately :-

Question

( A ) 3.4 eV
( B ) 1.9 eV
( C ) 1.5 eV
( D ) 0. 85 eV