Imagine that the electron in a hydrogen atom is replaced by a muon (μ). The mass of muon particle is 207 times that of an electron and charge is equal to the charge of an electron. The ionization potential of this hydrogen atom will be :

Solution

Physics sir · 3 steps

Bohr energy is proportional to mass, so a muon 207 times heavier is bound 207 times tighter.

1. Bohr energy formula. For a hydrogen-like atom, the energy of level n depends on the mass m of the orbiting particle. The charge is the same here, and the proton is treated as fixed, as the options expect.

   Eₙ = -(m e⁴)/(8 ε₀² h²) · Z²/n² ⇒ Eₙ ∝ m

2. Scale by mass. For ordinary hydrogen, the ionization energy from n = 1 is 13.6 eV. With m_μ = 207 mₑ, the ionization energy becomes 13.6 × 207 = 2815.2 eV.
3. So. The ionization potential of muonic hydrogen is 2815.2 V, so its ionization energy is 2815.2 eV. That is option (b).

The idea to keep. In Bohr's model energy goes as m Z²/n² and radius as n²/(mZ). Swap in a heavier particle with the same charge, and energy scales up while radius scales down by the mass ratio.

Answer from the NTA final answer key. Solution by PrepWiser.