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 :

Question

Aicharya · 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

A neutron moving with a speed 'v' makes a head on collision with a stationary hydrogen atom in ground state.…

Solution