Question

The radius of the first orbit of hydrogen is $r_H$ , and the energy in the ground state is $-13.6eV$ . Considering a ${\mu }^{-}$ -particle with a mass $207m_e$ revolving around a proton as in hydrogen atom, the energy and radius of proton and ${\mu }^{-}$ -combination respectively in the first orbit are (assume nucleus to be stationary)

• A. $-13.6\times 207eV,\frac{r_H}{207}$
• B. $-207\times 13.6eV,207r_H$
• C. $-\frac{13.6}{207}eV,\frac{r_H}{207}$
• D. $-\frac{13.6}{207}eV,207r_H$

Answer 1

Answer: (A)

The total energy of $n^{th}$ orbit,
$E_n=-\frac{m{e}^4}{8{ϵ}_0^2{h}^2}\cdot \frac{1}{n^2}$
Obviously $E_n\propto m$
$\therefore$ $\frac{E_{\mu }}{E_e}=\frac{m_{\mu }}{m_e}$
$\Rightarrow$ $E_{\mu }=\frac{m_{\mu }}{m_e}\times E_e$
Ground state energy of a proton in a hydrogen atom,
$E_{\mu }=-\text{13.6}\times \frac{207m_e}{m_e}eV$
$=-\text{13.6}\times 207eV$
( $\because m_{\mu }=207m_e,$ where $m_e$ is the mass of an electron)
We know that,
$r=\frac{{ϵ}_0{h}^2{n}^2}{207\pi m_e{e}^2}$
For ground state $\left(n=1\right)$ for proton, we have
$r_{\mu }=\frac{{ϵ}_0{h}^2}{207\pi m_e\cdot e^2}$
But $\frac{{ϵ}_0{h}^2}{\pi m_e\cdot e^2}$ = ground state radius of a hydrogen atom
$\frac{{ϵ}_0{h}^2}{\pi m_e\cdot e^2}=r_H$
$\therefore r_{\mu }=\frac{r_H}{207}$