Answer
$62nm$
Work Step by Step
(a) We can find the minimum uncertainty in position as follows:
$\frac{\Delta P}{P}=1.0\%=0.01$
$\implies \Delta P=(0.01)P$
$\implies \Delta P=(0.01)(1.7\times 10^{-25})=1.7\times 10^{-27}Kg.m/s$
Now $\Delta y=\frac{h}{2\pi \Delta P_y}$
We plug in the known values to obtain:
$\Delta y=\frac{6.63\times 10^{-34}}{(2)(3.14)(1.7\times 10^{-27})}$
$\Delta y=62\times 10^{-9}m=62nm$
