If the position of the electron is measured within an accuracy of $\pm 0.002 \,nm$,calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $\frac{h}{4 \pi \times 0.05 \,nm}$,is there any problem in defining this value?

From Heisenberg's uncertainty principle,$\Delta x \times \Delta p \geq \frac{h}{4 \pi}$.
Given,$\Delta x = 0.002 \,nm = 2 \times 10^{-12} \,m$.
$\Delta p = \frac{h}{4 \pi \Delta x} = \frac{6.626 \times 10^{-34} \,Js}{4 \times 3.1416 \times 2 \times 10^{-12} \,m} = 2.637 \times 10^{-23} \,kg \,ms^{-1}$.
The given momentum is $p = \frac{h}{4 \pi \times 0.05 \,nm} = \frac{6.626 \times 10^{-34} \,Js}{4 \times 3.1416 \times 5 \times 10^{-11} \,m} = 1.055 \times 10^{-24} \,kg \,ms^{-1}$.
Since the actual momentum $(1.055 \times 10^{-24} \,kg \,ms^{-1})$ is less than the uncertainty in momentum $(2.637 \times 10^{-23} \,kg \,ms^{-1})$,the value cannot be defined as it violates the uncertainty principle.