Question

The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 $\mathring{A}anditsgroundstateenergyequls-13.6eV.iftheelectroninthehydrogenatomisreplacedbymoon($\mu^{-}$)[chargesameaselectronandmass207m$_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)

$m_{\mu }=207m_e,q_{\mu }=q_{e^{-}},M_{\text{nucleus}}=1836m_e$
Reduced mass
$\mu =\frac{mM}{M+m}=\frac{207m_e\times 1836m_e}{207m_e+1836m_e}$
$=186m_e$
$\because r_1=\frac{n^2{h}^2}{4{\pi }^{2}mkz{e}^2}=0.51\mathring{A}$ (Given in Question)
Radius of first orbit of new atom
$r_1=\frac{m_e}{186m_e}\times 0.51\mathring{A}=2.56\times 10^{-13}m$
$E_1=\frac{\mu }{m}{E}_1=\frac{186m_e}{m_e}(-13.6eV)$
$=-2.8keV$