Question

A neutron of kinetic energy $81.6$ eV collides inelastically with a singly ionized helium atom at rest and is scattered at an angle of ${90}^{\circ }$. After scattering

• A. K.E of neutron can be 48.96 eV
• B. K.E of neutron can be 16.32 eV
• C. K.E of Helium can be 32.64 eV
• D. K.E of Helium can be 24.48 eV

Answer 1

Answer: (B), (D)

The correct options are
B K.E of neutron can be 16.32 eV
D K.E of Helium can be 24.48 eV


By COM $\Rightarrow m{V}_0=4m{V}_{2}cos\theta$ and $m{V}_1=4m{V}_{2}sin\theta$

$V_0^2+V_1^2=16V_2^2\dots \dots \left(1\right)$ where $\frac{1}{2}m{V}_0^2=81.6eV$

and $\frac{1}{2}m{V}_0^2=\frac{1}{2}m{V}_1^2+\frac{1}{2}4m{V}_2^2+Q\dots \left(2\right)$

where $Q=13.6\times 2^2(\frac{1}{1}-\frac{1}{n^2})$; n = 2, 3, 4, . . .
Consider the case where least amount of mechanical energy is lost, i.e., n = 2.
For n = 2 $\Rightarrow Q=40.8eV=\frac{1}{4}m{V}_0^2$

$\therefore$ By (2) $\Rightarrow \frac{1}{2}m{V}_0^2=\frac{1}{2}m{V}_1^2+\frac{1}{2}4m{V}_2^2+\frac{1}{4}m{V}_0^2$

$\Rightarrow \frac{V_0^2}{2}=V_1^2+4V_2^2\dots \dots \left(3\right)$

By (1) and (3) $\Rightarrow V_2^2=\frac{3}{40}{V}_0^2$ and $V_1^2=\frac{V_0^2}{5}$

$\frac{1}{2}m{V}_1^2=16.32eV$ and $\frac{1}{2}4m{V}_2^2=24.48eV$

So if there is a loss of mechanical energy, these are the maximum values of kinetic energies possible.