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 \(\mu\)-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.
Show Hint
For hydrogen-like systems,
\[
r_n\propto \frac{n^2}{Zm}.
\]
A heavier orbiting particle produces much smaller orbits.
Updated On: Jun 16, 2026
- \(10\)
- \(25\)
- \(104\)
- \(208\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For a hydrogen-like atom,
\[
r_n=\frac{n^2a_0}{Z}\frac{m_e}{m}
\]
where
\[
m=\text{mass of orbiting particle}.
\]
Step 1: Write the radius of the muonic atom. Given \[ Z=3 \] and \[ m=208\,m_e. \] Therefore, \[ r_n = \frac{n^2a_0}{3\times208}. \]
Step 2: Equate it to the first Bohr radius of hydrogen. For hydrogen first orbit, \[ r=a_0. \] Hence, \[ \frac{n^2a_0}{624} = a_0 \] \[ n^2=624 \] \[ n\approx24.98 \] \[ n\approx25. \] \[\begin{aligned} \boxed{25} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
Step 1: Write the radius of the muonic atom. Given \[ Z=3 \] and \[ m=208\,m_e. \] Therefore, \[ r_n = \frac{n^2a_0}{3\times208}. \]
Step 2: Equate it to the first Bohr radius of hydrogen. For hydrogen first orbit, \[ r=a_0. \] Hence, \[ \frac{n^2a_0}{624} = a_0 \] \[ n^2=624 \] \[ n\approx24.98 \] \[ n\approx25. \] \[\begin{aligned} \boxed{25} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.
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