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The expression for energy of the hydrogen atom for the ${\mathbf{n}}^{\mathbf{th}}$state is given by,

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$

The expression for the energy of the hydrogen atom for the lower excited state is given by,

${\mathbf{E}}_{\mathbf{1}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{1}}^{\mathbf{2}}}$ ….. (1)

The expression for the energy of the hydrogen atom for the higher excited state is given by,

${\mathbf{E}}_{\mathbf{2}}\mathbf{=}\left(-13.6\mathrm{eV}\right)\frac{\mathbf{1}}{{\mathbf{n}}_{\mathbf{2}}^{\mathbf{2}}}$ ….. (2)

Given that, the excitation energy is 10.2 eV and the ground state energy of the hydrogen atom is -13.6 eV. The initial energy of the hydrogen atom is equal to the sum of the energy of the ground state energy and the excitation energy.

$$\begin{gathered} {\mathrm{E}}_1=-13.6\mathrm{eV}+10.2\mathrm{eV} \\ =-3.4\mathrm{eV} \end{gathered}$$

The difference between the final energy of the atom and the initial energy of the atom energy is equal to the energy of the emitted photon.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{photon}}={\mathrm{E}}_2-{\mathrm{E}}_1 \\ =\left(-0.85\mathrm{eV}\right)-\left(-3.4\mathrm{eV}\right) \\ =2.55\mathrm{eV} \end{gathered}$$

Therefore, the energy of the photon emitted is 2.55.

Rearrange the equation (1) as below.

$n_1=\sqrt{-\frac{13.6eV}{E_1}}$

Substitute -0.85 eV for $E_1$in the above equation.

$$\begin{gathered} n_1=\sqrt{-\frac{13.6eV}{-0.85eV}} \\ =4 \end{gathered}$$

Therefore, the higher quantum number is 4.

Rearrange the equation (1).

$n_1=\sqrt{-\frac{13.6eV}{E_1}}$

Substitute -3.4 eV for ${\mathrm{E}}_1$in the above equation.

$$\begin{gathered} n_2=\sqrt{-\frac{13.6eV}{-3.4eV}} \\ =2 \end{gathered}$$

Hence, the lower quantum number is 2 .