Question

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $\nu_1$ and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $\nu_2$. The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that the positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is ___

Hint:

The ionization energy of positronium is half that of hydrogen. Energy conservation includes photon energies and kinetic energies of particles involved.

Answer 1

Step 1: Energy absorbed by hydrogen atom
The hydrogen atom absorbs a photon of energy \(h \nu_1\), which ejects the electron with kinetic energy 10 eV after ionization. The ionization energy of hydrogen is 13.6 eV, so: \[ h \nu_1 = 13.6\ \text{eV} + 10\ \text{eV} = 23.6\ \text{eV} \]
Step 2: Formation of positronium and photon emission
Electron and positron combine to form positronium in ground state, emitting a photon of energy \(h \nu_2\). The ionization energy of positronium is half that of hydrogen due to reduced mass effect: \[ E_{\text{pos}} = \frac{13.6}{2} = 6.8\ \text{eV} \]
Step 3: Kinetic energy of positronium atom
The positronium atom moves with kinetic energy 5 eV.
Step 4: Energy conservation
Total initial photon energy = energy of emitted photon + kinetic energies: \[ h \nu_1 = h \nu_2 + 5\ \text{eV} + 10\ \text{eV} \] Rearranged, the difference in photon energies is: \[ h \nu_1 - h \nu_2 = 15\ \text{eV} \]