Question

In previous question what is the minimum possible value of kinetic energy of the neutrons for this to be possible. The mass of neutron and proton can be assumed to be nearly same. Use $h c=12400\, eV\,\mathring{A}$.

• A. 31.875 eV
• B. 63.75 eV
• C. 127.5 eV
• D. 182.5 eV

Answer 1

Answer: (B)

Let $u$ be the speed of neutron before collision

At end of the deformation phase (when the kinetic energy of (neutron + $He^+$) system is least)

Where $u_{cm}$ is velocity of centre of mass. From conservation of momentum
$u_{ cm }=\frac{m u}{m+4 m}=\frac{u}{5}$
The loss of kinetic energy
$\Delta K=\frac{1}{2} m u^{2}-\frac{1}{2} n\left(\frac{u}{5}\right)^{2}-\frac{1}{2} 4 m\left(\frac{u}{5}\right)^{2}$
$\Rightarrow \Delta K=\frac{4}{5}\left(\frac{1}{2} m u^{2}\right)$
It $K$ is the kinetic energy of neutron then the maximum loss in K.E. of system is
$\frac{4}{5} K=12.75 \times 4=51\, eV$
or $K=\frac{51 \times 5}{4}=63.75\, eV$