Use the Bohr model result for hydrogen-like atoms

The energy of the $n$th level is $E_n = -13.6\,\text{eV}\times \frac{\mu}{m_e}\times \frac{Z^2}{n^2}$ where:

• $\mu$ is the mass of the orbiting particle if the nucleus is assumed very heavy,
• $m_e$ is the electron mass,
• $Z=1$ for hydrogen.

For ordinary hydrogen, the ionization potential from the ground state is $13.6\,\text{eV}$.

Replace electron by muon

Given: $m_\mu = 207\,m_e$

Since the muon has the same charge magnitude as the electron, the Coulomb attraction remains the same, but the energy scales directly with the particle mass.

Therefore, the ground-state binding energy becomes $E_1 = -13.6 \times 207\,\text{eV}$

Calculate the ionization potential

Ionization potential = magnitude of ground-state energy: $I = 13.6 \times 207$

$I = 2815.2\,\text{eV}$

Match with options

• A: $13.6\,\text{eV}$
• B: $2815.2\,\text{eV}$
• C: $331.2\,\text{eV}$
• D: $27.2\,\text{eV}$

Hence, the correct option is B.

Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.