Question

A \(\mu\)-meson of charge equal to that of an electron \((-e)\) and mass \(208\) times the mass of an electron moves in a circular orbit around a nucleus of charge \(+3e\). Assuming that the Bohr model is applicable and the mass of the nucleus is infinite, find the orbit number \(n\) for which the radius of the orbit is approximately the same as that of the first Bohr orbit of hydrogen atom.

• A. \(10\)
• B. \(25\)
• C. \(104\)
• D. \(208\)

Hint:

For hydrogen-like systems, \[ r_n\propto \frac{n^2}{Zm}. \] A heavier orbiting particle produces much smaller orbits.

Answer 1

The Correct Option is B

Step 1: Recall the Bohr radius for a hydrogen-like atom.
The orbit radius is $r_n = \frac{n^2 a_0}{Z}\cdot\frac{m_e}{m}$, where $a_0$ is the normal Bohr radius, $Z$ the nuclear charge number, and $m$ the mass of the orbiting particle.
Step 2: Put in the muon details.
Here the orbiting particle is a muon with $m = 208\,m_e$, and the nucleus has $Z = 3$. So $r_n = \frac{n^2 a_0}{3 \times 208}$.
Step 3: State the target.
We want this radius to equal the first Bohr radius of hydrogen, which is simply $a_0$.
Step 4: Set the two radii equal.
\[ \frac{n^2 a_0}{624} = a_0 \]
Step 5: Cancel $a_0$ and solve.
$n^2 = 624$, so $n = \sqrt{624} \approx 24.98$.
Step 6: Round to a whole orbit number.
Since the orbit number must be a whole number, $n \approx 25$.
\[ \boxed{n = 25} \]