Question

The speed of a certain proton is $4.5 \times 10^{5} ms ^{-1}$. If the uncertainty in its momentum is to be reduced to $0.010$ percent, what uncertainty in its location must be tolerated?

Answer 1

Mass of a proton $=1.67 \times 10^{-27} kg$

Momentum of proton $=m v=1.67 \times 10^{-27} \times 4.5 \times 10^{5}$

Uncertainty in momentum, $\Delta p$

$=1 \cdot 67 \times 10^{-27} \times 4 \cdot 5 \times 10^{5} \times 0 \cdot 01 \times 10^{-2}$

$=1 \cdot 67 \times 4.5 \times 10^{-26}$

$\therefore \Delta p \times \Delta x=\frac{h}{4 \pi} \\\therefore \Delta x=\frac{h}{4 \pi \Delta p}=\frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 1.67 \times 4.5 \times 10^{-26}} m$

$=0.070 \times 10^{-8} m =0.70 nm $