A neutron moving with a speed 'v' makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which perfactly inelastic collision

Question

A neutron moving with a speed 'v' makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which perfactly inelastic collision will take place is :

TestHub · Accepted Answer

Let velocity before collision v from momentum conservation $m v = (m + m) v^{1}$

$v^{1} = \frac{v}{2}$

Loss in K.E. $= \frac{1}{2} m v^{2} - \frac{1}{2} (2 m) {(\frac{v}{2})}^{2}$

$= \frac{1}{4} m v^{2}$

K.E. lost is used to jump from $1^{s t}$ orbit to $2^{n d}$ orbit $∆ K . E . = 10.2 e V$

$\Rightarrow \frac{1}{4} m v^{2} = 10.2$

$\frac{1}{2} m v^{2} = 2 \times 10.2 = 20.4 e V$

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