Answer
The kinetic energy of the daughter nucleus is $~~1.6\times 10^{-19}~J$
Work Step by Step
In part (a), we found that the magnitude of the momentum of the daughter nucleus is $~~p = 1.36\times 10^{-22}~kg~m/s$
We can find the speed of the daughter nucleus:
$m~v = p$
$v = \frac{p}{m}$
$v = \frac{1.36\times 10^{-22}~kg~m/s}{5.8\times 10^{-26}~kg}$
$v = 2345~m/s$
We can find the kinetic energy of the daughter nucleus:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(5.8\times 10^{-26}~kg)(2345~m/s)^2$
$K = 1.6\times 10^{-19}~J$
The kinetic energy of the daughter nucleus is $~~1.6\times 10^{-19}~J$
