Answer
(a) $K.E=0.6KeV$
(b) $K.E=1.5\times 10^{12}eV$
Work Step by Step
(a) We know that
$K.E=\frac{h^2}{2m\lambda^2}$
We plug in the known values to obtain:
$K.E=\frac{(6.634\times 10^{-34})^2}{2(9.11\times 10^{-31})(0.5\times 10^{-10})^2}(\frac{1eV}{1.60\times 10^{-19}})$
$K.E=0.6KeV$
(b) We know that
$K.E=\frac{h^2}{2m\lambda^2}$
We plug in the known values to obtain:
$K.E=\frac{(6.634\times 10^{-34})^2}{2(9.11\times 10^{-31})(10^{-15})^2}(\frac{1eV}{1.60\times 10^{-19}})$
$K.E=1.5\times 10^{12}eV$
