Answer
$12.26pm$
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:
$v=\sqrt{\frac{2\times 1.602\times 10^{-15}}{9.1095\times 10^{-31}}}=5.9306\times 10^7\frac{m}{s}$
We also know that
$\lambda=\frac{h}{mv}$
We plug in the known values to obtain:
$\lambda=\frac{6.626\times 10^{-34}}{9.1095\times 10^{-31}\times 5.9306\times 10^7}=1.226\times 10^{-11}m=12.26pm$