Question

A mu mesic atom is a hydrogen atom with electron replaced by muon whose mass is $210$ times of mass of electron, then –

• A. Bohr radius of the mumesic atom is $0.0023\stackrel{\circ }{\text{A}}$
• B. Ground state energy of mu mesic atom is $2856\text{eV}$
• C. Angular momentum in ground state is $\frac{h}{2\pi }$
• D. Angular momentum in ground state is $\frac{\left(210\right)h}{2}$

Answer 1

Answer: (A), (B), (C)

Angular momentum in ground state is $\frac{h}{2\pi }$

Radius of electron is given as :
$r=\frac{n^2{h}^2}{4{\pi }^{2}m{e}^{2}z}=0.529\frac{n^2}{z}{A}^{\circ }$
$r\propto \frac{1}{m}$
$\frac{r_{\mu }}{r_e}=\frac{m_e}{m_{\mu }}$
$\{\because m_{\mu }=210m_e\}$
$r_{\mu }=r_e\left(\frac{1}{210}\right)=0.53\times \frac{1}{210}=0.0023\stackrel{\circ }{\text{A}}$
Similarly, energy is given by
$E=-\frac{2{\pi }^{2}m{e}^4{z}^2}{n^2(4\pi {\epsilon }_0{)}^2}=-13.6\frac{z^2}{n^2}\text{eV}$
$E\propto m$
$E_{\mu }=E_e\left(\frac{m_{\mu }}{m_e}\right)=13.6\times 210=2856\text{eV}$
Angular momentum is given as
$L=mvr=\frac{nh}{2\pi }$
For $n=1$ : ground state
$L=\frac{h}{2\pi }$