Question

A neutron moving with a speed υ strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen = 1.67 × 10⁻²⁷ kg.

Answer 1

Given:
Mass of neutron, m = 1.67 × 10⁻²⁷ kg
Since neutron is moving with velocity $\left(v\right)$, its energy $\left(E\right)$ is given by
$E=\frac{1}{2}m{\upsilon }^2$
Let the energy absorbed be ∆E.
The condition for inelastic collision is given below:
$\frac{1}{2}m{\upsilon }^2>2∆E$
$\Rightarrow ∆E<\frac{1}{4}m{\upsilon }^2$

Since 10.2 eV, energy is required for the first excited state.
∴ ∆E < 10.2 eV
$\therefore 10.2\mathrm{eV}<\frac{1}{4}m{v}^2$
Thus, minimum speed of the neutron is given by
$$\begin{gathered} \Rightarrow {\upsilon }_{\min }=\sqrt{\frac{4\times 10.2}{m}}\mathrm{eV} \\ \Rightarrow {\upsilon }_{\min }=\sqrt{\frac{10.2\times 1.6\times {10}^{19}\times 4}{1.67\times {10}^{-27}}} \\ =6\times {10}^4\mathrm{m}/\sec \end{gathered}$$