To excite the spectral line of wavelength $4960 \stackrel{\circ }{\mathrm{A}}$ of an atom, an excitation energy of $7.7 \mathrm{eV}$ is required. The ground state energy of the atom is $10.5 \mathrm{eV}$. The energies of two levels involved in the emission of $4960 \stackrel{\circ }{\mathrm{A}}$ line are (Assume he $=1240 \mathrm{eV} \mathrm{nm}$, where $h$ is Planck's constant and $\mathrm{c}$ is speed of light)

• A

  $\begin{array}{ll}14.2\mathrm{eV},& 16.1\mathrm{eV}\end{array}$
• B

  $12.2\mathrm{eV},18.2\mathrm{eV}$
• C

  $15.7\mathrm{eV},20.5\mathrm{eV}$
• Correct answer
  D

  $15.7\mathrm{eV}, 18.2\mathrm{eV}$

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