Answer
$\Delta x\ge5.54\times10^{-10}m$
Work Step by Step
$KE=\frac{1}{2}mv^2=\frac{1}{2}m\frac{m}{m}v^2=\frac{p^2}{2m}$
$p=\sqrt{2mKE}=3.8202\times10^{-23}kgm/s$
$p'=\sqrt{2mKE1.01}=3.8392\times10^{-23}kgm/s$
$\Delta p=1.905\times10^{-25}kgm/s$
$\Delta x\ge\frac{\hbar}{\Delta p}$
$\Delta x\ge5.54\times10^{-10}m$
