Question

A neutron moving witha speed v strikes a hydrogen atom in ground state towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completly or partially) collision may take place. The mass of neutron = mass of hydrogen = $1.67\times {10}^{-27}kg$

Answer 1

Energy of the neutron is $\frac{1}{2}m{v}^2$

The condition for inelastic collision is

$\frac{1}{2}m{v}^2>2\Delta E$

$\Rightarrow \Delta E<\frac{1}{4}m{v}^2$

Here $\Delta E$ is the energy absorted.

Energy required for first excited state is 10.2 e V.

$\therefore \Delta E<10.2eV$

$\therefore 10.2eV<\frac{1}{4}m{V}^2$

$\Rightarrow v_{m/n}=\sqrt{\frac{4\times 10.2}{m}}eV$

$\Rightarrow v=\sqrt{\frac{10.2\times 1.6\times {10}^{19}\times 4}{1.67\times {10}^{-27}}}$

= $6\times {10}^{-27}$ m/sec