Answer
a) $v=0.6427c$
b) $P=2.29\times 10^{-22}Kg.m/s$
c) $P=2.29\times 10^{-22}Kg.m/s$
d) $v=9.85Km/s$
Work Step by Step
(a) We know that
$\frac{v}{c}=\sqrt{1-\frac{1}{(1+\frac{K}{m_{\circ}c^2})^2}}$
We plug in the known values to obtain:
$\frac{v}{c}=\sqrt{1-\frac{1}{(1+\frac{156KeV}{511KeV})^2}}$
This simplifies to:
$v=0.6427c$
(b) We can find the required momentum as follows:
$P=\frac{m_{\circ}v}{\sqrt{1-\frac{v^2}{c^2}}}$
We plug in the known values to obtain:
$P=\frac{(9.11\times 10^{-31}Kg)(0.6427c)}{\sqrt{1-(\frac{0.6427c}{c})^2}}$
$\implies P=2.29\times 10^{-22}Kg.m/s$
(c) We know that
$P_e+P_{\beta}=0$
$\implies P_e=-P_{\beta}$
$\implies P_e=-(2.29\times 10^{-22}Kg.m/s)$
(d) We know that
$v=c[1+\frac{m_{\circ}^2c^2}{p^2}]^{-\frac{1}{2}}$
We plug in the known values to obtain:
$v=(3\times 10^8m/s)[1+\frac{(14.003242)^2(1.66\times 10^{-27}Kg)^2(3\times 10^8m/s)^2}{(-2.29\times 10^{-22}Kg.m/s)^2}]^{-\frac{1}{2}}$
$v=9.85Km/s$