A muon of charge -eand mass$\mathit{m}\mathbf{=}\mathbf{207}{\mathit{m}}_{\mathbf{e}}$(where${\mathit{m}}_{\mathbf{e}}$is the mass of an electron) orbits the nucleus of a singly ionized helium atom (He+). Assuming that the Bohr model of the hydrogen atom can be applied to this muon–helium system, verify that the energy levels of the system are given by

$\mathit{E}\mathbf{=}\mathbf{-}\frac{\mathbf{11}\mathbf{.}\mathbf{3}\mathbf{e}\mathbf{V}}{{\mathbf{n}}^{\mathbf{2}}}$

The rotational equivalent of linear momentum is called angular momentum. The formula for angular momentum is given by,

L = mvr ……. (1)

Where,L is angular momentum, m mass of object, v is speed of object and r is distance.

Here we have, mass $m=2071m_e=207\left(1.6\times {10}^{-18}C\right)$

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

We set the central force to be equal to the mass times the radial central acceleration because we first need to determine the permitted values of the radius. In other words:

$\frac{m{v}^2}{r}=\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}$…… (2)

$v=\sqrt{\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{mr}}}$

Where m is the mass of helium, v is the velocity of moun, r is a distance of the moun from the center of the Helium atom, and ${\epsilon }_0$is permittivity.

Now, put the value ofinequation (1), we get,

$$\begin{gathered} L=m\sqrt{\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0\mathrm{mr}}r} \\ L=\sqrt{\frac{Z{e}^{2}mr}{4{\mathrm{\pi \epsilon }}_0}}.....................\left(3\right) \end{gathered}$$

Now, also, we know that $L=\frac{nh}{2\mathrm{\pi }}$

So, substitute value of L in equation (3) we get,

$$\begin{gathered} \frac{nh}{2\mathrm{\pi }}=\sqrt{\frac{Z{e}^{2}m{r}_n}{4{\mathrm{\pi \epsilon }}_0}} \\ {\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{h}}^{2}4{\mathrm{\epsilon }}_0}{4{\mathrm{\pi Ze}}^2\mathrm{m}} \end{gathered}$$

Now, the total energy is given by,

$$\begin{gathered} E=K+U \\ =\frac{1}{2}m{v}^2-\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0}.\frac{1}{r_n} \\ =\frac{Z{e}^2}{8{\mathrm{\pi \epsilon }}_0}.\frac{1}{r}-\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0}.\frac{1}{r}\left(\because byequation\left(2\right)\right) \\ =-\frac{Z{e}^2}{4{\mathrm{\pi \epsilon }}_0}.\frac{1}{r_n} \end{gathered}$$

Now, by substituting the value of $r_n$we get,

$$\begin{gathered} E_n=-\frac{Z{e}^2}{8{\mathrm{\pi \epsilon }}_0}.\left(\frac{4{\mathrm{\pi Ze}}^2\mathrm{m}}{n^2{h}^{2}4{\epsilon }_0}\right) \\ =-\frac{Z^2{e}^{4}m}{8{{\epsilon }_0}^2{h}^2}.\frac{1}{n^2} \end{gathered}$$

Now, substitute all the numerical values in above equation. So we get,

$$\begin{gathered} E_n=-\frac{{\left(2\right)}^2{\left(1.6\times {10}^{-18}C\right)}^4\left(207\left(1.6\times {10}^{-18}C\right)\right)}{8{\left(8.85\times {10}^{-12}F/m^2\right)}^2{\left(6.63\times {10}^{-34}J.s\right)}^2}.\frac{1}{n^2} \\ =-\frac{1.08\times {10}^{-15}J}{n^2} \end{gathered}$$

Now, we know that$1eV=1.6\times {10}^{-19}J$

So, the value of becomes

$$\begin{gathered} E_n=-\frac{1.80\times {10}^{-15}J}{\left(1.6\times {10}^{-19}J/eV\right)n^2} \\ =\frac{-1.13\times {10}^{4}eV}{n^2} \\ =-\frac{-11.3keV}{n^2} \end{gathered}$$

Hence, it is verify that the energy levels of the system are given by$E_n=-\frac{-11.3keV}{n^2}$ .