Answer
a) $1.07nJ$
b) $0.151nJ$
c) $0.92nJ$
Work Step by Step
(a) We know that
$E=\frac{m_{\circ}c^2}{\sqrt{1-\frac{v^2}{c^2}}}$
We plug in the known values to obtain:
$E=\frac{(1.675\times 10^{-27}Kg)(3\times 10^8m/s)^2}{\sqrt{1-(\frac{0.99c}{c})^2}}$
$E=1.07\times 10^{-9}J=1.07nJ$
(b) We can find the rest energy of the neutron as follows:
$m_{\circ}c^2=(1.675\times 10^{-27}Kg)(3\times 10^8m/s)^2=0.151nJ$
(c) The kinetic energy of the neutron is as follows:
$K.E=\frac{m_{\circ}c^2}{\sqrt{1-\frac{v^2}{c^2}}}-m_{\circ}c^2$
We plug in the known values to obtain:
$K.E=1.07nJ-0.151nJ$
$K.E=0.92nJ$
