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

• A

  $1.5$ $\mathrm{e}\mathrm{V}$
• B

  $0.85$ $\mathrm{e}\mathrm{V}$
• C

  $3.4$ $\mathrm{e}\mathrm{V}$
• Correct answer
  D

  $1.9$ $\mathrm{e}\mathrm{V}$

Hence, the correct answer is option: 'D'.