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JEE Advanced 2025Paper 2
Exam
JEE Advanced
Year
2025
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2025 · Paper 2 · Physics · Atomic Physics

A hydrogen atom, initially at rest in its ground state, absorbs a...

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v1v_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 v2v_2. The center of mass of the resulting positronium atom moves with a kinetic energy of 5 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 \qquadA hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v_1 and ejects the electron with

Correct answer

11.8

Step-by-step solution

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency v_1 and ejects the electron with
hv1=13.6+10=23.6eV....(1)\mathrm{h} v_1=13.6+10=23.6 \mathrm{eV} \qquad ....(1) Energy of positronium in ground state =−13.6μm(zn)2eV=−13.6×12eV=−6.8eV\begin{aligned} & =-13.6 \frac{\mu}{m}\left(\frac{z}{n}\right)^2 \mathrm{eV} \\ & =-13.6 \times \frac{1}{2} \mathrm{eV}=-6.8 \mathrm{eV}\end{aligned} So to make positronium 6.8 eV must release \& 5 eV is the KE of COM . So total energy of photon released (hv2)\left(h v_2\right) will be : hv2=(10−5)+6.8=11.8eV....(2)∴\mathrm{h} v_2=(10-5)+6.8=11.8 \mathrm{eV} \qquad ....(2) \therefore \quad Difference in energy =23.6−11.8=11.8eV=23.6-11.8=11.8 \mathrm{eV}
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