Question

In a process a neutron which is initially at rest, decays into a proton, an electron and an antineutrino. The ejected electron has a momentum of $p_1=2.4\times {10}^{-26}kgm/s$ and the antineutrino has $p_2=7.0\times {10}^{-27}kgm/s$ . Find the recoil speed of the proton if the electron and the antineutrino are ejected (a) along the same direction (b) in mutually perpendicular directions (Mass of the proton $m_p=1.67\times {10}^{-27}$ )

Answer 1

(a)

Let the velocity of the proton is $V_p$.

The momentum is given as,

$1.67\times {10}^{-27}\times V_p=2.4\times {10}^{-26}+7\times {10}^{-27}$

$V_p=18.56m/s$

Thus, the recoil speed of proton is $18.56m/s$.

(b)

The total momentum of electron and antineutrino is given as,

$p=\sqrt{{\left(24\right)}^2+{\left(7\right)}^2}\times {10}^{-27}$

$p=25\times {10}^{-27}kgm/s$

The recoil speed of proton is given as,

$1.67\times {10}^{-27}\times V_p=25\times {10}^{-27}$

$V_p=14.97m/s$

Thus, the recoil speed of proton is $14.97m/s$.