Solution: Photon Energy from Protonium Transition
Key Concept: Protonium is an exotic atom consisting of a proton and an antiproton. The energy levels of hydrogen-like atoms depend linearly on the reduced mass ($\mu$) of the system.
$$ E_n \propto \mu $$
Step 1: Calculate Reduced Mass ($\mu$)
Mass of proton $m_p \approx 1836 m_e$. For Protonium ($p + \bar{p}$): $$ \mu_{Pp} = \frac{m_p \cdot m_p}{m_p + m_p} = \frac{m_p}{2} $$ For Hydrogen ($p + e$): $$ \mu_{H} \approx m_e $$ Ratio of reduced masses: $$ \frac{\mu_{Pp}}{\mu_{H}} = \frac{m_p/2}{m_e} \approx \frac{1836}{2} = 918 $$
Mass of proton $m_p \approx 1836 m_e$. For Protonium ($p + \bar{p}$): $$ \mu_{Pp} = \frac{m_p \cdot m_p}{m_p + m_p} = \frac{m_p}{2} $$ For Hydrogen ($p + e$): $$ \mu_{H} \approx m_e $$ Ratio of reduced masses: $$ \frac{\mu_{Pp}}{\mu_{H}} = \frac{m_p/2}{m_e} \approx \frac{1836}{2} = 918 $$
Step 2: Scale the Energy
The energy of a transition in Hydrogen ($2 \to 1$) is: $$ \Delta E_H = 13.6 \text{ eV} \times \left( \frac{1}{1^2} – \frac{1}{2^2} \right) = 13.6 \times 0.75 = 10.2 \text{ eV} $$ The energy for Protonium will be scaled by the mass ratio 918: $$ \Delta E_{Pp} = 918 \times 10.2 \text{ eV} $$ $$ \Delta E_{Pp} \approx 9363.6 \text{ eV} \approx 9.36 \text{ keV} $$
The energy of a transition in Hydrogen ($2 \to 1$) is: $$ \Delta E_H = 13.6 \text{ eV} \times \left( \frac{1}{1^2} – \frac{1}{2^2} \right) = 13.6 \times 0.75 = 10.2 \text{ eV} $$ The energy for Protonium will be scaled by the mass ratio 918: $$ \Delta E_{Pp} = 918 \times 10.2 \text{ eV} $$ $$ \Delta E_{Pp} \approx 9363.6 \text{ eV} \approx 9.36 \text{ keV} $$
Correct Option: (b) 9.37 keV