Q156 / 160 · AP EAMCET 2024 (20 May Shift 1) · 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
    $\mathrm{n} \approx 20$
  • Correct answer
    B
    $\mathrm{n} \approx 25$
  • C
    $\mathrm{n} \approx 28$
  • D
    $\mathrm{n} \approx 29$

Solution

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 get $\frac{h^2}{4 \pi^2 \mathrm{mke}^2}=\frac{\mathrm{n}^2 \mathrm{~h}^2}{208 \times 3 \times 4 \pi^2 \mathrm{mke}^2}$ $\Rightarrow \mathrm{n}=\sqrt{208 \times 3} \approx 25$

Hence, the correct answer is option: 'B'.

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