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(A) muon is an unstable elementary particle of mass nearly $200 \, m_{e}$ and charge $\pm e$.Here,a negative muon is bound to a proton.So,$m = 200 \,

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$X$-rays

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(A) muon is an unstable elementary particle of mass nearly $200 \, m_{e}$ and charge $\pm e$.Here,a negative muon is bound to a proton.So,$m = 200 \, m_{e}$ and $M = 1836 \, m_{e}$ (as the mass of a proton is $1836$ times the mass of an electron).The reduced mass of the system is,$m^{\prime} = \frac{m M}{m + M} = \frac{200 \, m_{e} 	imes 1836 \, m_{e}}{200 \, m_{e} + 1836 \, m_{e}} \approx 180 \, m_{e}$.As the mass of the muon is comparable to the mass of the proton,we must account for the motion of the nucleus by calculating the reduced mass.The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_{n} = \frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}$.Thus,the energy of a muon in a muonic atom is $E_{n}^{\prime} = \frac{m^{\prime} e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}} = 180 \, E_{n}$.Considering a Balmer transition from $n = 3$ to $n = 2$,the energy difference is $\Delta E^{\prime} = E_{n=3}^{\prime} - E_{n=2}^{\prime} = \frac{hc}{\lambda}$.$\Delta E^{\prime} = 180 	imes (E_{n=3} - E_{n=2}) = 180 	imes \left( \frac{-13.6 \, 	ext{eV}}{3^{2}} - \frac{-13.6 \, 	ext{eV}}{2^{2}} ight) = 180 	imes 1.89 \, 	ext{eV} = 340.2 \, 	ext{eV}$.Using $\lambda = \frac{hc}{\Delta E^{\prime}} = \frac{1240 \, 	ext{eV} \cdot 	ext{nm}}{340.2 \, 	ext{eV}} \approx 3.65 \, 	ext{nm}$.This wavelength falls in the range of $X$-rays ($0.01 \, 	ext{nm}$ to $10 \, 	ext{nm}$).

In a muonic atom,a muon of mass $200$ times that of an electron and the same charge is bound to a proton. The wavelengths of its Balmer series are in the range of

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