# Solve the following :

Question:

A neutron initially at rest decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of $1.4 \times 10^{-26} \mathrm{Kg}-\mathrm{m} / \mathrm{s}$ and the antineutrino $6.4 \times 10^{-27} \mathrm{Kg}-\mathrm{m} / \mathrm{s}$. Find the recoil speed of the proton (a) if the electron and the antineutrino are ejected along the same direction and (b) if they are ejected along perpendicular directions. Mass of the proton $=1.67 \times 10^{-27} \mathrm{Kg}$.

Solution:

$\mathrm{n} \rightarrow p+e^{-}+\bar{v}$

$p_{e^{-}}=1.410^{-26} \mathrm{Kg}-\mathrm{m} / \mathrm{s}$

$p_{v^{-}}=6.4 \times 10^{-27} \mathrm{Kg}-\mathrm{m} / \mathrm{s}$

(a) If $e^{-}$and ${ }^{\bar{v}}$ are along same direction.

Resultant velocity, $V_{r}=\sqrt{V_{e}^{-2}+V_{v}^{-2}+2 V_{e} V_{\bar{v}} \cos 0^{\circ}}$

$=\left(V_{\bar{e}}+V_{\bar{v}}\right)$ in same direction.

C.O.L.M

$\Longrightarrow 0=p_{e}^{-}+p_{\bar{v}}+p_{p}$

$\Rightarrow V_{p}=\frac{-\left(p_{\bar{p}}^{-}+p_{\bar{v}}\right)}{m_{p}}=-\frac{2.04 \times 10^{-26}}{1.67 \times 10^{-27}}=12.25 \mathrm{~m} / \mathrm{s}$

(b) If $\theta=90^{\circ}, V_{r}=\sqrt{V_{e^{-}}^{2}+V_{\bar{v}}^{2}}$ or $p_{r}=\sqrt{p_{e^{-}}^{2}+p_{\bar{v}}^{2}}$ $p_{r}+p_{p}=0$

$V_{p}=\frac{-p_{r}}{m_{p}}=\frac{-\sqrt{(0.64)^{2}+(1.4)^{2}} \times 10^{-13}}{1.67 \times 10^{-27}}$

$=9.19 \sim 9.2 \mathrm{~m} / \mathrm{s}$