Question

An electron has an uncertainty in its position of 471 pm .

Answer 1

Answer:

$\Delta x . \Delta p \geq \frac{h}{4 \pi}$

∆x is the Uncertainty in the position of the particle

∆p is the Uncertainty in the momentum of the particle

$\Delta p=\frac{h}{4 \pi \times m \times \Delta x}$

$=\frac{6.626 \times 10^{-34} \mathrm{kgm}^{2} s^{-1}}{4 \times 3.14 \times 9.109 \times 10^{-31} \mathrm{kg} \times 471 \times 10^{-12} \mathrm{m}}$

$=\frac{6.626 \times 10^{-34} \mathrm{ms}^{-1}}{53887 \times 10^{-43}}$

$=1.23\times10^5 ms^{-1}$ is the answer