Question

The radius of the first orbit of hydrogen is $r_{H}$ , and the energy in the ground state is $-13.6\,eV$ . 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. $-207\times 13.6 \, eV, \, 207r_{H}$
• 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,
$E_{n}= \, -\frac{me^{4}}{8 \epsilon _{0}^{2} h^{2}}\cdot \frac{1}{n^{2}}$
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,
$E_{\mu }=-\text{13.6}\times \frac{207 m_{e}}{m_{e}} \, eV$
$=-\text{13.6}\times 207 \, eV$
( $\because \,m_{\mu }=207m_{e},$ where $m_{e}$ is the mass of an electron)
We know that,
$r=\frac{\epsilon _{0} h^{2} n^{2}}{207 \pi m_{e} e^{2}}$
For ground state $\left(n= 1\right)$ for proton, we have
$r_{\mu }=\frac{\epsilon _{0} h^{2}}{207 \pi m_{e} \cdot e^{2}}$
But $\frac{\epsilon _{0} h^{2}}{\pi m_{e} \cdot e^{2}}$ = ground state radius of a hydrogen atom
$\frac{\epsilon _{0} h^{2}}{\pi m_{e} \cdot e^{2}}=r_{H}$
$\therefore r_{\mu }=\frac{r_{H}}{207}$