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 the mass 207`m_e` revolving round a porton as in hydrogen atom, the energy and radius of proton and `mu`-combination respectively in the first orbit are(assume nucleus to be stationary)

Options

A. `-14.6xx207eV,r_(H)/207`
B. `-207xx13.6 eV,207r_H`
C. `-(13.6)/(207)eV.(r_H)/207`
D. `-(13.6)/(207)eV,207r_H`

Answer

a

Explanation

proton is,
`r_mu=(epsilon _0h^2)/(207pim_ee^2)
r_mu=(r_H)/207[therefore r_H=(epsilon_0h^2)/(pim_ee^2)]`