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 $207$ $m_e$ revolving round 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. $-207\times 13.6eV$, $207$ $r_H$
• C. $\frac{-13.6}{207}eV$, $\frac{r_H}{207}$
• D. $\frac{-13.6}{207}eV$, $207$ $r_H$

Answer 1

The correct option is A $-13.6\times 207eV$, $\frac{r_H}{207}$

Radius of ground state, $r=\frac{h^2}{4{\pi }^{2}mK{e}^2}$

$⟹$ $r\propto \frac{1}{m}$

$⟹$ $\frac{r_{\mu }}{r_H}=\frac{m_e}{m_{\mu }}=\frac{m_e}{207m_e}$

$⟹$ $r_{\mu }=\frac{r_H}{207}$

Energy of ground state, $E=-\frac{K{e}^2}{2r}$

$⟹$ $E\propto m$ $(\because r\propto \frac{1}{m})$

$⟹$ $\frac{E_{\mu }}{-13.6}=\frac{m_{\mu }}{m_e}=\frac{207m_e}{m_e}$

$⟹$ $E_{\mu }=-13.6\times 207$ $eV$