Question

The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 $\mathring{A} and its ground state energy equls -13.6 eV. if the electron in the hydrogen atom is replaced by moon ($\mu^{-}$ ) [charge same as electron and mass 207 m$_e$], the first Bohr radius and ground state energy will be :

• A. 0.53 × 10$^{-13}$ m, -3.6 eV
• B. 25.6 × 10$^{-13}$ m, -2.8 eV
• C. 2.56 × 10$^{-13}$ m, -2.8 eV
• D. 2.56 × 10$^{-13}$ m, -13.6 eV

Answer 1

Answer: (C)

$m_{\mu} = 207\,m_e, q_{\mu} = q_{e^-}, M_{\text{nucleus}} = 1836\,m_e$
Reduced mass
$\mu = \frac{mM}{M + m} = \frac{207 m_e \times 1836 m_e}{207 m_e + 1836 m_e}$
$ = 186\,m_e$
$\because r_1 = \frac{n^2h^2}{4\pi^2 mkze^2} = 0.51\,\mathring{A}$ (Given in Question)
Radius of first orbit of new atom
$r_1 = \frac{m_e}{186m_e} \times 0.51\mathring{A} = 2.56\times 10^{-13} m$
$E_1 = \frac{\mu}{m}E_1 = \frac{186\,m_e}{m_e} (-13.6 \,eV)$
$ = -2.8\,keV$