Question

A \( \mu \)-meson of charge \( e \), mass \( 208 \, m_e \) moves in a circular orbit around a heavy nucleus having charge \( +3e \). The quantum state \( n \) for which the radius of the orbit is the same as that of the first Bohr orbit for the hydrogen atom is (approximately):

• A. \( n \approx 20 \)
• B. \( n \approx 25 \)
• C. \( n \approx 28 \)
• D. \( n \approx 29 \)

Hint:

For a hydrogen-like system with a different mass or nuclear charge, use the modified Bohr radius formula and solve for \( n \). The key factors affecting the radius are the mass of the orbiting particle and the nuclear charge \( Z \).

Answer 1

The Correct Option is B

Step 1: Understanding Bohr Radius Formula The radius of the \( n \)th orbit in a hydrogen-like atom is given by: \[ r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} \] where: - \( h \) is Planck’s constant, - \( \varepsilon_0 \) is permittivity of free space, - \( m \) is the mass of the orbiting particle, - \( Z \) is the atomic number (here, \( Z = 3 \)), - \( e \) is the charge of an electron, - \( n \) is the quantum number. For hydrogen, the first Bohr radius is: \[ r_1 = \frac{h^2 \varepsilon_0}{\pi m_e e^2} \] Step 2: Finding \( n \) for the Muon System For the muon system, the radius equation modifies due to the different mass and nuclear charge: \[ r_n^{\mu} = \frac{n^2 h^2 \varepsilon_0}{\pi (208 m_e) (3e^2)} \] Setting this equal to the first Bohr radius of hydrogen: \[ \frac{n^2 h^2 \varepsilon_0}{\pi (208 m_e) (3e^2)} = \frac{h^2 \varepsilon_0}{\pi m_e e^2} \] Step 3: Solving for \( n \) Dividing both sides by \( \frac{h^2 \varepsilon_0}{\pi m_e e^2} \): \[ \frac{n^2}{208 \times 3} = 1 \] \[ n^2 = 208 \times 3 \] \[ n^2 = 624 \] \[ n \approx \sqrt{624} \approx 25 \] Step 4: Conclusion Thus, the quantum number for which the muon’s orbit matches the first Bohr radius of hydrogen is \( n \approx 25 \).