A free hydrogen atom in ground state is at rest. A neutron of kinetic energy ‘K’ collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy 2.55 eV. (Assume that the hydrogen atom and neutron has same mass)
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Question

A free hydrogen atom in ground state is at rest. A neutron of kinetic energy ‘K’ collides with the hydrogen atom. After collision hydrogen atom emits two photons in succession one of which has energy 2.55 eV. (Assume that the hydrogen atom and neutron has same mass)

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G1 Physics · Accepted Answer

Step 1: For a hydrogen-like atom, Eₙ = −13.6Z²/n² eV and photon energy is ΔE = 13.6Z²|1/n₂²−1/n₁²| = hc/λ. Step 2: Use the stated transition/excitation condition to obtain the required value. Step 3: Final: A: minimum value of ‘K’ is 25.5 eV.; C: the other photon has energy 10.2 eV if K is minimum.; D: the upper energy level is of excitation energy 12.75 eV. 20.