Question

A neutron moving with some kinetic energy collides head-on with a stationary helium ion $\left(H{e}^{+}\right)$ in its ground state. The minimum kinetic energy of neutron so that the collision can be inelastic, is:

• A. $54.4eV$
• B. $32.6eV$
• C. $51eV$
• D. $40.8eV$

Answer 1

The correct option is D $40.8eV$

Collision between neutron and $H{e}^{+}$ will be inelastic if a part of kinetic energy is used to excite ion if $u_1$ and $u_2$ are speed of neutron and $H{e}^{+}$
$mu=m{u}_1+m{u}_2$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{u}^2+\frac{1}{2}\left(2m\right)u_2^2+OE$
$u^2=u_1^2+2(u-4{)}^2+\frac{2\Delta E}{m}$
${24}^2-u{u}_1+\frac{2\Delta E}{m}=0$ for $u_1$ to be
${24}^2-\frac{4\Delta E}{m}\ge 0$
$\frac{m{v}^2}{2}\ge 4\Delta E$
$\Delta E=10.2eV$
$=40.8eV$