Question

Anti muonic atom is with anti-muon (200 times massive to electron and having -ve electronic charge) and a positron. Ground state energy of hydrogen atom is -13.6 eV. The energy of atom in 9th excited state in anti-muonic atom is-

• A. -$\frac{272}{2010}$ eV
• B. -13.6 eV
• C. -6.8 eV
• D. -0.06 eV

Answer 1

Answer: (A)

A

Answer 2

The energy levels of a hydrogen-like atom are given by $E_n = -\frac{Z^2 \mu k^2 e^4}{2 \hbar^2 n^2}$. For hydrogen, the ground state energy ($n=1$) is $E_{1,H} \approx -\frac{m_e k^2 e^4}{2 \hbar^2} = -13.6$ eV.

For the anti-muonic atom, the anti-muon has mass $m_{a\mu} = 200 m_e$ and the positron has mass $m_p = m_e$. The reduced mass is $\mu_{anti-\mu} = \frac{m_e (200 m_e)}{m_e + 200 m_e} = \frac{200}{201} m_e$.

The energy levels can be expressed as $E_n \approx \frac{\mu_{anti-\mu}}{m_e} \left(\frac{-13.6 \text{ eV}}{n^2}\right)$. The 9th excited state corresponds to $n = 10$. $E_{10} \approx \frac{\frac{200}{201} m_e}{m_e} \left(\frac{-13.6 \text{ eV}}{10^2}\right) = \frac{200}{201} \left(\frac{-13.6 \text{ eV}}{100}\right) = \frac{200}{20100} (-13.6 \text{ eV}) = \frac{2}{201} (-13.6 \text{ eV}) = -\frac{27.2}{201}$ eV. This value is equivalent to -$\frac{272}{2010}$ eV.