If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron.

From Heisenbero's uncertainty principle.

$\Delta x \times \Delta p=\frac{\mathrm{h}}{4 \pi} \Rightarrow \Delta p=\frac{1}{\Delta x} \cdot \frac{\mathrm{h}}{4 \pi}$

Where,

$\Delta x=$ uncertainty in position of the electron

$\Delta p=$ uncertainty in momentum of the electron

Substituting the values in the expression of $\Delta p:$

$\Delta p=\frac{1}{0.002 \mathrm{~nm}} \times \frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times(3.14)}$

$=\frac{1}{2 \times 10^{-12} \mathrm{~m}} \times \frac{6.626 \times 10^{-34} \mathrm{JS}}{4 \times 3.14}$

$=2.637 \times 10^{-23} \mathrm{~J} \mathrm{sm}^{-1}$

$\Delta p=2.637 \times 10^{-23} \mathrm{kgms}^{-1}\left(1 \mathrm{~J}=1 \mathrm{kgms}^{2} \mathrm{~s}^{-1}\right)$

$\therefore$ Uncertainty in the momentum of the electron $=2.637 \times 10^{-23} \mathrm{kgms}^{-1}$.

Actual momentum $=\frac{\mathrm{h}}{4 \pi_{\mathrm{m}} \times 0.05 \mathrm{~nm}}$

$=\frac{6.626 \times 10^{-34} \mathrm{Js}}{4 \times 3.14 \times 5.0 \times 10^{-11} \mathrm{~m}}$

$=1.055 \times 10^{-24} \mathrm{kgms}^{-1}$

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.