Question

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

• A. X-rays
• B. infrared rays
• C. $\gamma$-rays
• D. microwave

Answer 1

Answer: (A)

A muon is an unstable elementary particle of mass nearly $200m_e$ and charge $\pm e$
Here, a negative muon is given bound to a proton.

So, $m=200m_e$ and $M=1836m_e$
(as mass of a proton is $1836$ times of mass of electron)
So, reduced mass of system, $m^{\prime }=\frac{mM}{m+M}=\frac{200m_e\times 1836m_e}{200m_e+1836m_e}\approx 180m_e$
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,
$E_n=\frac{m{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$
So, energy of a muon of muonic atom is
$E_n^{\prime }=\frac{m^{\prime }{e}^4}{8{\epsilon }_0^2{h}^2{n}^2}$
$\Rightarrow E_n^{\prime }=180E_n$
Now, considering a Balmer transition, $n=3$ to $n=2$.
$\therefore \Delta E^{\prime }=\frac{hc}{\lambda }$
or $\Delta E^{\prime }=E_{n=3}^{\prime }-E_{n=2}^{\prime }=\frac{hc}{\lambda }$
$\Rightarrow 180\left(E_{n=3}-E_{n=2}\right)=\frac{hc}{\lambda }$
$\Rightarrow 180\left(\frac{-13.6}{3^2}-\frac{-13.6}{2^2}\right)=\frac{hc}{\lambda }$
$\Rightarrow 180\times 1.89eV=\frac{1240(eV-nm)}{\lambda (nm)}$
$\Rightarrow \lambda (nm)=3.6nm$
This emitted radiation is X-rays.
Note X-rays have a wavelength range of $0.01nm$ to $10nm$.