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(A) In a head-on elastic collision between a neutron of mass $m$ and a hydrogen atom of mass $m$ (approximately),the velocity of the center of mass is

Vedclass · Accepted Answer

$20.4$

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(A) In a head-on elastic collision between a neutron of mass $m$ and a hydrogen atom of mass $m$ (approximately),the velocity of the center of mass is $v_{cm} = v/2$.After the collision,both particles move with the same velocity $v/2$ in the center of mass frame.The loss in kinetic energy is given by $\Delta K = K_{initial} - K_{final} = \frac{1}{2}mv^2 - [\frac{1}{2}m(v/2)^2 + \frac{1}{2}m(v/2)^2] = \frac{1}{2}mv^2 - \frac{1}{4}mv^2 = \frac{1}{4}mv^2$.For an inelastic collision to occur,this lost kinetic energy must be at least equal to the excitation energy of the hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$,which is $\Delta E = 10.2 \ eV$.Therefore,$\frac{1}{4}mv^2 = 10.2 \ eV$.The initial kinetic energy of the neutron is $K = \frac{1}{2}mv^2 = 2 	imes (\frac{1}{4}mv^2) = 2 	imes 10.2 \ eV = 20.4 \ eV$.

$A$ neutron moving with a speed $v$ makes a head-on collision with a stationary hydrogen atom in its ground state. The minimum kinetic energy of the neutron for which an inelastic collision will take place is....$eV$

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