Question

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

Answer 1

From Hisenberg'g uncertainity principle;

$\Delta x.\Delta p\ge \frac{h}{4\pi }\Delta x=\pm 0.02\times {10}^{-9}m=2\times {10}^{-12}m\therefore \Delta p\ge \frac{6.64\times {10}^{-34}}{4\times 3.14}\times \frac{{10}^{12}}{2}=0.264\times {10}^{-22}Kgm{s}^{-1}$

So, uncertainity in momentum is $\pm 0.264\times {10}^{-22}kgm{s}^{-1}$

We can't take exact value of momentum because it will violate Hisenberg's uncertainity principle .So, we can't take $p=\frac{h}{4\pi m}$.