Physics - Modern Physics Question with Solution | TestHub

Key Idea Hydrogen atom can be considered to be the system of two charges, positive charged nucleus and negative charged electron. A system of this kind is equivalent to a single particle of mass $m^{\prime }$ that revolves around the position of the heavier particle. Then, the reduced mass of electron is $$m^{\prime }=\frac{mM}{m+M}$$ where, $m=$ mass of electron and $$M=\text{ mass of nucleus }$$ Its values is less than $m$. Given, radius of first orbit for electron ${\mathrm{r}}_1=0.51\mathring{A}$, ground state energy of electron, ${\mathrm{E}}_1=-13.6\mathrm{eV}$, mass of electron $={\mathrm{m}}_{\mathrm{e}}$ mass of muon, ${\mathrm{m}}_{\mu }=207{\mathrm{m}}_{\mathrm{e}}$ and mass of nucleus, $M=1836{\mathrm{m}}_{\mathrm{e}}$ When electron in hydrogen atom is replaced by muon, the reduced mass of muon is $${\mathrm{m}}_{\mu }^{\prime }=\frac{{\mathrm{m}}_{\mu }\mathrm{M}}{{\mathrm{m}}_{\mu }+\mathrm{M}}$$ Substituting the given values in Eq. (i), we get $${\mathrm{m}}_{\mu }^{\prime }=\frac{207{\mathrm{m}}_{\mathrm{e}}\times 1836{\mathrm{m}}_{\mathrm{e}}}{207{\mathrm{m}}_{\mathrm{e}}+1836{\mathrm{m}}_{\mathrm{e}}}\approx 186{\mathrm{m}}_{\mathrm{e}}$$ The radius of first orbit in hydrogen atom for electron is given by, $${\mathrm{r}}_1=\frac{{\mathrm{h}}^2{\epsilon }_0}{\pi {\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^2}$$ The radius of first orbit for muon is $${\mathrm{r}}_1^{\prime }=\frac{{\mathrm{h}}^2{\epsilon }_0}{\pi {\mathrm{m}}_{\mu }^{\prime }{\mathrm{e}}^2}$$ $$\begin{array}{rl}& \left[\because \text{ charge of }\mu =\text{ charge of }{\mathrm{e}}^{-}\right]\\ & =\frac{{\mathrm{h}}^2{\epsilon }_0}{\pi \times 186{\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^2}\\ & =\left(\frac{{\mathrm{h}}^2{\epsilon }_0}{\pi {\mathrm{m}}_{\mathrm{e}}{\mathrm{e}}^2}\right)\frac{\mathrm{l}}{186}=\frac{{\mathrm{r}}_1}{186}\\ & =\frac{0.51}{186}\mathring{A}\left[\because {\mathrm{r}}_1=0.51\mathring{A}\right]\\ & =2.74\times 10^{-13}\mathrm{m}\left[\because 1\mathring{A}=10^{-10}\mathrm{m}\right]\\ & \end{array}$$ The total energy of electron is given by $${\mathrm{E}}_{\mathrm{n}}=\frac{-{\mathrm{mZ}}^2{\mathrm{e}}^4}{8{\epsilon }_0^2{\mathrm{h}}^2}\left(\frac{1}{{\mathrm{n}}^2}\right)\Rightarrow {\mathrm{E}}_{\mathrm{n}}\propto \mathrm{m}$$ For electron in first orbit of hydrogen atom, $${\mathrm{E}}_1={\mathrm{km}}_{\mathrm{e}}$$ where, $\mathrm{k}=\frac{{\mathrm{e}}^4}{8{\epsilon }_0^2{\mathrm{h}}^2}=$ constant. For muon in first orbit $$\begin{array}{rlr}{\mathrm{E}}_1{}^{\prime }& ={\mathrm{km}}_{\mu }{}^{\prime }& \\ & =\mathrm{k}\times 186{\mathrm{m}}_{\mathrm{e}}& \text{ [from Eq. (i)] }\\ & =186{\mathrm{km}}_{\mathrm{e}}& \\ & =186{\mathrm{E}}_1& \text{ [from Eq. (iv)] }\\ & =186(-13.6)\mathrm{eV}& \text{ (given) }\\ & =-2529.6\mathrm{eV}& \\ & =-2.5\mathrm{keV}& \end{array}$$ $\therefore$ The values are closest to that of options (3).