Question

A neutron with an energy of $4.6\text{MeV}$ collides with protons and is retarded. Assuming that upon each collision the neutron is deflected by ${45}^{\circ }$ The number of collisions which will reduce its energy to $0.23\text{eV}.$ nearly is (integer only.)

Answer 1

Given, initial K.E of the neutron is $K_0=4.6\text{MeV};\theta ={45}^{\circ }$

Mass of neutron $\approx$ Mass of proton =m

From conservation of momentum in $y$-direction

$\sqrt{2m{K}_1}\sin {45}^{\circ }=\sqrt{2m{K}_2}\sin \theta ....\left(1\right)$

Similarly, $x$-direction $\sqrt{2m{K}_0}-\sqrt{2m{K}_1}\cos {45}^{\circ }=\sqrt{\sqrt{2m{K}_2}cos\theta }.....\left(2\right)$

$\text{Squaring and adding equation (1) and (2), We have}$

$K_2=K_1+K_0-\sqrt{2K_0{K}_1}......\left(3\right)$

$\text{From conservation of energy}$

$K_2=K_0-K_1.....\left(4\right)$

$\text{Solving equations (3) and (4), we get}$

$K_1=\frac{K_0}{2}$

$\text{i.e., after each collision energy remains half. Therefore, after n colliisions,}$

$K_n=K_0{\left(\frac{1}{2}\right)}^n$

$\therefore 0.23=(4.6\times {10}^6){\left(\frac{1}{2}\right)}^n$

$\Rightarrow 2^n=\frac{4.6\times {10}^6}{0.23}$

$\Rightarrow n\log 2=\log (2\times {10}^7)=\log 2+7$

$\Rightarrow n=1+\frac{7}{\log 2}=1+\frac{7}{0.3}\approx 24$