Answer
$K.E=1.07\times 10^{-17}J$
Work Step by Step
We can find the kinetic energy of the electron as follows:
$\lambda=\frac{h}{p}$
This can be rearranged as:
$p=\frac{h}{\lambda}$
We plug in the known values to obtain:
$p=\frac{6.63\times 10^{-34}}{1.5\times 10^{-10}}=4.42\times 10^{-24}Kg\frac{m}{s}$
Now $K.E=\frac{p^2}{2m}$
We plug in the known values to obtain:
$K.E=\frac{(4.42\times 10^{-24})^2}{2\times 9.1\times 10^{-31}}$
$K.E=1.07\times 10^{-17}J$
