Answer
The uncertainty in position of the proton is $\Delta x\ge0.315nm$.
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 proton: $m\approx1.673\times10^{-27}kg$
- The uncertainty in velocity: $\Delta v=0.01\times10^{4}m/s$
Since we only have the uncertainty in velocity, the uncertainty in momentum $\Delta(mv)=m\Delta v$
2) Calculate the uncertainty in position
$\Delta x\ge\frac{h}{4\pi m\Delta v}=\frac{6.626\times10^{-34}}{4\pi\times(1.673\times10^{-27})\times(0.01\times10^4)}\approx3.152\times10^{-10}m\approx0.315nm$
