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Question: A neutron having kinetic energy (K), collides head-on with a stationary H-atom. Find the nature of c...

A neutron having kinetic energy (K), collides head-on with a stationary H-atom. Find the nature of collision (elastic/inelastic/perfectly inelastic), when the neutron possesses a kinetic energy of: (i) 12 eV (ii) 20.4 eV (iii) 22 eV (iv) 24.18 eV

A

12 eV: Elastic; 20.4 eV: Inelastic; 22 eV: Inelastic; 24.18 eV: Inelastic

B

12 eV: Inelastic; 20.4 eV: Elastic; 22 eV: Elastic; 24.18 eV: Elastic

C

12 eV: Elastic; 20.4 eV: Elastic; 22 eV: Inelastic; 24.18 eV: Inelastic

D

12 eV: Inelastic; 20.4 eV: Inelastic; 22 eV: Elastic; 24.18 eV: Elastic

Answer

12 eV: Elastic; 20.4 eV: Inelastic; 22 eV: Inelastic; 24.18 eV: Inelastic

Explanation

Solution

The minimum energy required to excite a hydrogen atom from its ground state (n=1) to the first excited state (n=2) is ΔE=E2E1=3.4 eV(13.6 eV)=10.2 eV\Delta E = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}.

For a head-on collision between a neutron and a stationary hydrogen atom of approximately equal masses, the collision becomes inelastic if the neutron's initial kinetic energy (KiK_i) is sufficient to cause excitation. The threshold kinetic energy (KthK_{th}) for an inelastic collision is given by Kth=2×ΔEminK_{th} = 2 \times \Delta E_{min}.

Using ΔEmin=10.2\Delta E_{min} = 10.2 eV: Kth=2×10.2 eV=20.4 eVK_{th} = 2 \times 10.2 \text{ eV} = 20.4 \text{ eV}.

If K<KthK < K_{th}, the collision is elastic. If KKthK \ge K_{th}, the collision is inelastic.

(i) Kinetic energy of 12 eV: 12 eV<20.4 eV    12 \text{ eV} < 20.4 \text{ eV} \implies Elastic collision. (ii) Kinetic energy of 20.4 eV: 20.4 eV=20.4 eV    20.4 \text{ eV} = 20.4 \text{ eV} \implies Inelastic collision. (iii) Kinetic energy of 22 eV: 22 eV>20.4 eV    22 \text{ eV} > 20.4 \text{ eV} \implies Inelastic collision. (iv) Kinetic energy of 24.18 eV: 24.18 eV>20.4 eV    24.18 \text{ eV} > 20.4 \text{ eV} \implies Inelastic collision.