A neutron collides head-on with a stationary | vedclass

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Question

(A) For an inelastic collision to occur,the neutron must transfer at least the minimum excitation energy of the hydrogen atom,which is $\Delta E = 10.

Vedclass · Accepted Answer

If the kinetic energy of the neutron is less than $10.2\,eV,$ the collision must be elastic.

Vedclass · Answer

(A) For an inelastic collision to occur,the neutron must transfer at least the minimum excitation energy of the hydrogen atom,which is $\Delta E = 10.2\,eV$ (transition from $n=1$ to $n=2$).Let $m$ be the mass of the neutron and the hydrogen atom (approximately equal). Let $v$ be the initial velocity of the neutron,and $v_1, v_2$ be the velocities of the neutron and hydrogen atom after the collision.Conservation of momentum: $mv = mv_1 + mv_2 \implies v = v_1 + v_2$.Conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \Delta E$.Substituting $v_1 = v - v_2$ into the energy equation:$\frac{1}{2}mv^2 = \frac{1}{2}m(v - v_2)^2 + \frac{1}{2}mv_2^2 + \Delta E$$\frac{1}{2}mv^2 = \frac{1}{2}m(v^2 - 2vv_2 + v_2^2) + \frac{1}{2}mv_2^2 + \Delta E$$0 = -mvv_2 + mv_2^2 + \Delta E \implies mv_2^2 - mvv_2 + \Delta E = 0$.For $v_2$ to be real,the discriminant $D = (-mv)^2 - 4(m)(\Delta E) \geq 0$.$m^2v^2 \geq 4m\Delta E \implies \frac{1}{2}mv^2 \geq 2\Delta E$.Given $\Delta E = 10.2\,eV,$ the minimum kinetic energy required for an inelastic collision is $K_{min} = 2 	imes 10.2\,eV = 20.4\,eV.$ If $K < 10.2\,eV,$ the collision must be elastic.

$A$ neutron collides head-on with a stationary hydrogen atom in its ground state.

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