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Q.
In a muonic atom, a muon of mass of $200$ times of that of electron and same charge is bound to the proton. The wavelengths of its Balmer series are in the range of
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Solution:
A muon is an unstable elementary particle of mass nearly $200 m_{e}$ and charge $\pm e$
Here, a negative muon is given bound to a proton.
So, $m=200 \,m_{e}$ and $M=1836\, m _{e}$
(as mass of a proton is $1836$ times of mass of electron)
So, reduced mass of system, $m^{\prime}=\frac{m M}{m+M}=\frac{200\, m_{e} \times 1836 \,m_{e}}{200 m_{e}+1836 \,m_{e}} \approx 180\, m_{e}$
As mass of muon is comparable to mass of proton,
we have to take account of motion of nucleus.
That's way we are calculating reduced mass.
Now, as energy of an electron in $n$th orbit is of $H$-atom,
$E_{n}=\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}$
So, energy of a muon of muonic atom is
$E_{n}'=\frac{m' e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}$
$ \Rightarrow E_{n}'=180 E_{n}$
Now, considering a Balmer transition, $n=3$ to $n=2$.
$\therefore \Delta E'=\frac{h c}{\lambda}$
or $\Delta E'=E_{n=3}'-E_{n=2}'=\frac{h c}{\lambda} $
$\Rightarrow 180\left(E_{n=3}-E_{n=2}\right)=\frac{h c}{\lambda}$
$\Rightarrow 180\left(\frac{-13.6}{3^{2}}-\frac{-13.6}{2^{2}}\right)=\frac{h c}{\lambda}$
$\Rightarrow 180 \times 1.89\, eV =\frac{1240( eV - nm )}{\lambda( nm )} $
$\Rightarrow \lambda( nm )=3.6 \,nm$
This emitted radiation is X-rays. Note X-rays have a wavelength range of $0.01 \,nm$ to $10\ nm$.
