91Ó°ÊÓ

Energyof an atom in the nth level of the hydrogen atom is given by,

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{e}}{\mathbf{e}}^{\mathbf{4}}}{\mathbf{8}{{\mathbf{Î\mu }}_{\mathbf{0}}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}\mathbf{.}\frac{\mathbf{1}}{{\mathbf{n}}^{\mathbf{2}}}$……. (1)

Where,${\mathit{E}}_{\mathbf{n}}$is energy produce by electron, ${\mathit{m}}_{\mathbf{e}}$is mass of electron, e is electric charge of electron, ${\mathit{Î\mu }}_{\mathbf{0}}$is permittivity, h is plank’s constant and n is Quantum number.

From equation (1) we have,Energy of an atom in the nth level of the hydrogen atom is given by,

$E_n=\frac{m_e{e}^4}{8{{\mathrm{Î\mu }}_0}^2{h}^2}.\frac{1}{n^2}$

Now, we have,$m_e=9.11×{10}^{-31}\mathrm{kg}$

$$\begin{gathered} e=1.6×{10}^{-18}\mathrm{C} \\ {\mathrm{Î\mu }}_0=8.85×{10}^{-12}\mathrm{F}/{\mathrm{m}}^2 \\ h=6.63×{10}^{-34}\mathrm{J}.\mathrm{s} \end{gathered}$$

Now, put all values in equation (1) we get,

$$\begin{gathered} E_n=\frac{(9.11×{10}^{-31}\mathrm{kg}){(1.6×{10}^{-18}\mathrm{C})}^4}{8{(8.85×{10}^{-12}\mathrm{F}/{\mathrm{m}}^2)}^2{\left(6.63×10-34\mathrm{J}.\mathrm{s}\right)}^2}.\frac{1}{{\mathrm{n}}^2} \\ =\frac{2.18×{10}^{-18}}{{\mathrm{n}}^2} \end{gathered}$$

Now, we know thatrole="math" localid="1661838773953" $1\mathrm{eV}=1.6×{10}^{-19}\mathrm{J}$

By substituting $1\mathrm{eV}=1.6×{10}^{-19}\mathrm{J}$in above equation, we get

$$\begin{gathered} E_n=\frac{2.18×{10}^{-18}}{n^2(1.6×{10}^{-19}J/eV)}J \\ =-\frac{13.6eV}{n^2} \end{gathered}$$

Now, in equation 39-33 we get, $E_n=-\frac{13.61eV}{n^2}$.

So, we can say thatthat the combined value of the constants appearing in Eq. 39-33 is 13.6 eV .

Hence it is verified.