Question

The radius of the first permitted Bohr orbit for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals -13.6 eV. If the electron in the hydrogen atom is replaced by muon [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 keV
• D. 2.56 × $10^{-13}$ m, -13.6 eV

Answer 1

Answer: (C)

C

Answer 2

The Bohr radius $r_n$ is inversely proportional to the mass of the orbiting particle ($r_n \propto 1/m$). The ground state energy $E_1$ is directly proportional to the mass of the orbiting particle ($E_1 \propto m$).

Given the first Bohr radius for electron $r_e = 0.529 \text{ Å}$ (using the standard value for consistency with options) and ground state energy $E_e = -13.6 \text{ eV}$. The muon mass is $m_\mu = 207 m_e$.

1. First Bohr Radius for Muon ($r_\mu$): $r_\mu = r_e \times \frac{m_e}{m_\mu} = 0.529 \text{ Å} \times \frac{m_e}{207 m_e} = \frac{0.529}{207} \text{ Å} \approx 0.0025555 \text{ Å} = 2.56 \times 10^{-13} \text{ m}$.

2. Ground State Energy for Muon ($E_\mu$): $E_\mu = E_e \times \frac{m_\mu}{m_e} = -13.6 \text{ eV} \times \frac{207 m_e}{m_e} = -13.6 \times 207 \text{ eV} = -2815.2 \text{ eV} = -2.8152 \text{ keV} \approx -2.8 \text{ keV}$.

Comparing these results with the options, option (C) is the correct match.