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

To solve this problem, we need to compare the radius of the \( \mu \)-meson’s orbit around a heavy nucleus to the first Bohr orbit of hydrogen. The radius of the \( n \)-th Bohr orbit is given by:

\[ r_n = \frac{n^2 h^2}{4 \pi^2 k e^2 m} \]

For hydrogen, the radius of the first Bohr orbit (\( n=1 \)) is:

\[ r_1 = \frac{h^2}{4 \pi^2 k e^2 m_e} \]

Here, \( m_e \) is the electron mass. For the \( \mu \)-meson orbiting a heavy nucleus with charge \( +3e \), the effective nuclear charge is \( Ze = 3e \). The Bohr radius for the \( \mu \)-meson is:

\[ r_n' = \frac{n^2 h^2}{4 \pi^2 k (3e)e m_\mu} \]

Given \( m_\mu = 208 m_e \), the formula becomes:

\[ r_n' = \frac{n^2 h^2}{12 \pi^2 k e^2 \cdot 208 m_e} \]

We need \( r_n' = r_1 \) for the radius of the first Bohr orbit, hence:

\[ \frac{n^2 h^2}{12 \pi^2 k e^2 \cdot 208 m_e} = \frac{h^2}{4 \pi^2 k e^2 m_e} \]

Simplifying, we find:

\[ \frac{n^2}{2496} = 1 \]

The value of \( n^2 \) is approximately 2496, so:

\[ n \approx \sqrt{2496} \approx 50 \]

This was an error in calculation. Correctly solving the simplified equation after considering multiplicative constants gives:

\[ \frac{n^2}{624} \approx 1 \Rightarrow n^2 \approx 624 \Rightarrow n \approx 25 \]

The correct answer is therefore approximately \( n \approx 25 \).

Answer 2

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 \( n \approx 25 \).