AP EAMCET · PHYSICS · Atomic Physics
\(\mu\)-meson of charge ' \(e\) ', mass 208 me moves in a circular orbit around a heavy nucleus having charge \(+3 e\). The quantum state ' \(n\) ' for which the radius of the orbit is same as that of the first Bohr orbit for hydrogen atom ts [approximately]
- A \(\mathrm{n} \approx 20\)
- B \(\mathrm{n} \approx 25\)
- C \(\mathrm{n} \approx 28\)
- D \(\mathrm{n} \approx 29\)
Answer & Solution
Correct Answer
(B) \(\mathrm{n} \approx 25\)
Step-by-step Solution
Detailed explanation
For hydrogen atom, \(r_n=\frac{n^2 h^2}{4 \pi^2 m k e^2} \Rightarrow r_1=\frac{h^2}{4 \pi^2 m k e^2}\) ....(i) For \(\mu\)-meson, \(\mathrm{m} \rightarrow 208 \mathrm{~m}, \mathrm{Z}=3\) \(\therefore r_n=\frac{n^2 h^2}{4 \pi^2(208 m) k(3) e^2}\) ....(ii) From eq(i) and (ii), we…
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