Answer
$19.4pm$
Work Step by Step
We know that
$E_k=\frac{1}{2}mv^2$
This can be rearranged as
$v=\sqrt{\frac{2E_k}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{2\times 6.408\times 10^{-16}}{9.1095\times 10^{-31}}}=3.7508\times 10^7\frac{m}{s}$
We also know that
$\lambda=\frac{h}{mv}$
We plug in the known values to obtain:
$\lambda=\frac{6.626\times 10^{-34}}{9.1095\times 10^{-31}\times 3.7508\times 10^7}=1.94\times 10^{-11}m=19.4pm$
