JEE Challenger
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PhyHardAdv 2025 • Paper 2

Difference Between Photon Energies in Hydrogen Ionization and Positronium Formation

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency ν1\nu_1 and ejects the electron with a kinetic energy of 10 eV10\text{ 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 ν2\nu_2. The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV5\text{ eV}. It is given that 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 ____

Official Numerical Answer11.7 to 11.9

Topics & Concepts

AtomsBohr Model

Step-by-Step Solution

To find the difference between the energies of the two photons, we analyze each step of the process using energy conservation and the Bohr model of the atom.

Step 1: Energy of the first photon (hν1h\nu_1)

A hydrogen atom in its ground state has an ionization energy of EI=13.6 eVE_I = 13.6\text{ eV}. When it absorbs a photon of frequency ν1\nu_1 and ejects an electron with kinetic energy Ke=10 eVK_e = 10\text{ eV}, the energy of the absorbed photon is given by: hν1=EI+Keh\nu_1 = E_I + K_e hν1=13.6 eV+10 eV=23.6 eVh\nu_1 = 13.6\text{ eV} + 10\text{ eV} = 23.6\text{ eV}

Step 2: Ground state energy of the positronium atom

A positronium atom consists of an electron and a positron orbiting their common center of mass. Since both particles have equal mass mem_e, the reduced mass μ\mu of the positronium system is: μ=me⋅meme+me=me2\mu = \frac{m_e \cdot m_e}{m_e + m_e} = \frac{m_e}{2}

In the Bohr model, the energy levels are directly proportional to the reduced mass of the system: En∝μE_n \propto \mu

Since the ground state energy of hydrogen (where μ≈me\mu \approx m_e) is −13.6 eV-13.6\text{ eV}, the ground state energy of positronium (EPs,1E_{\text{Ps}, 1}) is: EPs,1=12×(−13.6 eV)=−6.8 eVE_{\text{Ps}, 1} = \frac{1}{2} \times (-13.6\text{ eV}) = -6.8\text{ eV}

Step 3: Energy of the second photon (hν2h\nu_2)

The electron (kinetic energy Ke=10 eVK_e = 10\text{ eV}) and the positron at rest (Kp=0 eVK_p = 0\text{ eV}) combine to form a positronium atom moving with center of mass kinetic energy KCM=5 eVK_{\text{CM}} = 5\text{ eV} and emit a photon of frequency ν2\nu_2.

By the principle of conservation of energy: Einitial=EfinalE_{\text{initial}} = E_{\text{final}} Ke+Kp=EPs,1+KCM+hν2K_e + K_p = E_{\text{Ps}, 1} + K_{\text{CM}} + h\nu_2

Substituting the known values: 10 eV+0 eV=−6.8 eV+5 eV+hν210\text{ eV} + 0\text{ eV} = -6.8\text{ eV} + 5\text{ eV} + h\nu_2 10 eV=−1.8 eV+hν210\text{ eV} = -1.8\text{ eV} + h\nu_2 hν2=10 eV+1.8 eV=11.8 eVh\nu_2 = 10\text{ eV} + 1.8\text{ eV} = 11.8\text{ eV}

Step 4: Difference between the two photon energies

The difference between the energies of the two photons is: ΔE=hν1−hν2\Delta E = h\nu_1 - h\nu_2 ΔE=23.6 eV−11.8 eV=11.8 eV\Delta E = 23.6\text{ eV} - 11.8\text{ eV} = 11.8\text{ eV}

Thus, the difference between the two photon energies is 11.8.

Difference Between Photon Energies in Hydrogen Ionization and Positronium Formation | Physics PYQ Solution - JEE Challenger