Question

A neutron initially at rest,decays into a proton,an electron and an antineutrino. The ejected electron has a momentum of $1.4\times {10}^{-26}kg-\frac{m}{s}$ and the antinuetrino $6.4\times {10}^{-27}kg-\frac{m}{s}$.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 =$1.67\times {10}^{-27}kg$.

Answer 1

Mass of proton=$1.67\times {10}^{-27}kg$

Let $"V_p"$ be the velocity of proton.

Given,moment of electron= $1.4\times {10}^{-26}kg-\frac{m}{s}$

and momentum of antineutrino=$6.4\times {10}^{-27}kg-\frac{m}{s}$

(a) The electron and the antineutrino are ejected in the same direction.

As the total momentum is conserved,the proton should be ejected in the opposite direction.

$1.67\times {10}^{-27}\times V_p$

=$1.4\times {10}^{-26}+6.4\times {10}^{-27}$

=$20.4\times {10}^{-27}$

$\therefore V_p=\left(\frac{20.4}{1.67}\right)$

=12.2 m/sec in the opposite direction

(b) The electron and antineutrino are ejected perpendicular to each other.

Total momentum of electron and antineutrino

$=\sqrt{(14{)}^2+(6.4{)}^2}\times {10}^{-27}kg-\frac{m}{s}$

$\Rightarrow V_p=\frac{15.4\times {10}^{-27}kg-\frac{m}{sec}}{1.67\times {10}^{-27}kg}$

$\therefore V_p=9.2\frac{m}{s}$