Answer
The uncertainty in position of the electron is $\Delta x\ge57.89nm$.
Work Step by Step
*The Heisenberg's uncertainty principle formula: $$\Delta x\times\Delta(mv)\ge\frac{h}{4\pi}$$
$\Delta x$: the uncertainty in position
$\Delta(mv)$: the uncertainty in momentum
$h$: Planck's constant ($h=6.626\times10^{-34}J.s$)
1) Find the known variables
- Mass of electron: $m\approx9.109\times10^{-31}kg$
- The uncertainty in velocity: $\Delta v=0.01\times10^{5}m/s$
Since we only have the uncertainty in velocity, the uncertainty in momentum $\Delta(mv)=m\Delta v$
2) Calculate the uncertainty in position of the electron
$\Delta x\ge\frac{h}{4\pi m\Delta v}=\frac{6.626\times10^{-34}}{4\pi\times(9.109\times10^{-31})\times(0.01\times10^5)}\approx5.789\times10^{-8}m\approx57.89nm$
