Question:
A neutron moving with a speed \(v\) makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of neutron for which inelastic collision will take place is
A neutron moving with a speed \(v\) makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of neutron for which inelastic collision will take place is
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For a head-on collision between equal masses, only half the kinetic energy of the projectile is available in the centre-of-mass frame. Hence the threshold is twice the excitation energy.
Updated On: Apr 20, 2026
- 10.2 eV
- 20.4 eV
- 12.1 eV
- 16.8 eV
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
For inelastic collision with hydrogen atom: neutron must supply at least excitation energy (10.2 eV). In lab frame, not all KE is available for excitation.
Step 2: Detailed Explanation:
Using conservation of momentum and energy. Masses are equal (neutron \(\approx\) proton). Condition: \(\frac{1}{2}mv^2 - \frac{1}{2}(v_1 - v_2)^2 \geq \Delta E\). Threshold condition: \(\frac{1}{2}mv_{\min}^2 = 2\Delta E\). Therefore: \(KE_{\min} = 2 \times 10.2 = 20.4\text{ eV}\).
Step 3: Final Answer:
Minimum kinetic energy \(= 20.4\) eV.
For inelastic collision with hydrogen atom: neutron must supply at least excitation energy (10.2 eV). In lab frame, not all KE is available for excitation.
Step 2: Detailed Explanation:
Using conservation of momentum and energy. Masses are equal (neutron \(\approx\) proton). Condition: \(\frac{1}{2}mv^2 - \frac{1}{2}(v_1 - v_2)^2 \geq \Delta E\). Threshold condition: \(\frac{1}{2}mv_{\min}^2 = 2\Delta E\). Therefore: \(KE_{\min} = 2 \times 10.2 = 20.4\text{ eV}\).
Step 3: Final Answer:
Minimum kinetic energy \(= 20.4\) eV.
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