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

Question

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

MARKS App · Accepted Answer

$\Delta x=0.002 \mathrm{~nm}=2 \times 10^{-3} \mathrm{~nm}=2 \times 10^{-12} \mathrm{~m}$ $\begin{aligned} \Delta x \times \Delta p &=\frac{\mathrm{h}}{4 \pi} \ \therefore \Delta p &=\frac{\mathrm{h}}{4 \pi \Delta x}=\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times\left(2 \times 10^{-12} \mathrm{~m}\right)} \ &=2.638 \times 10^{-23} \mathrm{~kg} \mathrm{~ms}^{-1} \end{aligned}$ Actual momentum $\begin{aligned} &=\frac{\mathrm{h}}{4 \pi \times 0.05 \mathrm{~nm}}=\frac{\mathrm{h}}{4 \pi \times 5 \times 10^{-11} \mathrm{~m}} \ &=\frac{6.626 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-1}}{4 \times 3.14 \times 5 \times 10^{-11} \mathrm{~m}} \ &=1.055 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1} \end{aligned}$ This value cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty in momentum, which is impossible.

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