Question

If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value.

Answer 1

From Heisenberg’s uncertainty principle,
$△x\times △p=\frac{h}{4\pi }\Rightarrow △p=\frac{1}{△x}.\frac{h}{4\pi }$
Where,
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:
$△p=\frac{1}{0.002nm}\times \frac{6.626\times {10}^{-34}Js}{4\times \left(3.14\right)}=\frac{1}{2\times {10}^{-12}m}\times \frac{6.626\times {10}^{-34}Js}{4\times 3.14}=2.637\times {10}^{-23}Js{m}^{-1}△p=2.637\times {10}^{-23}kgm{s}^{-1}(1J=1kgm{s}^2{s}^{-1})$
$\therefore$ Uncertainty in the momentum of the electron = 2.637 × ${10}^{–23}$ kgm$s^{–1}$.
Actual momentum $=\frac{h}{4{\pi }_m\times 0.05nm}=\frac{6.626\times {10}^{-34Js}}{4\times 3.14\times 5.0\times {10}^{-11}m}=1.055\times {10}^{-24}kgm{s}^{-1}$
Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.