For muonic hydrogen, a muon (having mass 207 times the mass of electron) orbits around proton (having mass 1836 times the mass of electron). Approximated orbital radius in ground state for such system is A×10−B in m. Here, A is integer lesser than 10. Find B−2A.

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For muonic hydrogen, a muon (having mass 207 times the mass of electron) orbits around proton (having mass 1836 times the mass of electron). Approximated orbital radius in ground state for such system is $A \times 10^{- B}$ in m. Here, A is integer lesser than 10. Find $B - 2 A .$

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Detailed Solution

Here, $m = 207 m_{e}$ and $M = 1836 m_{e}$ So the reduced mass is, $m^{'} = \frac{m M}{m + M}$
$= \frac{(207 m_{e}) (1836 m_{e})}{207 m_{e} + 1836 m_{e}} = 186 m_{e}$
The orbit radius corresponding to $n = 1$ is $r_{1} = \frac{h^{2} ε_{0}}{π m_{e} e^{2}}$
Where, $r_{1} = a_{0} = 5.29 \times 10^{- 11} m$
Hence, the radius $r^{'}$ that correspond to the reduced mass $m^{'}$ is $r_{1}^{'} = (\frac{m}{m^{'}}) r_{1} = (\frac{m}{186 m_{e}}) a_{0}$
$= 2.85 \times 10^{- 13} \approx 3 \times 10^{- 13} m \Rightarrow A = 3 a n d B = 13 \Rightarrow \frac{B}{A} = \frac{13}{3} = 4.33$