Question

The uncertainty in the position of electron ($\Delta x$) is approximately 100 pm. The uncertainty in momentum (in kg m s$^{-1}$) of an electron is [h = 6.626 $\times 10^{-34}$ Js]

• A. $1.104 \times 10^{-22}$
• B. $0.527 \times 10^{-27}$
• C. $0.527 \times 10^{-24}$
• D. $1.055 \times 10^{-24}$

Hint:

Remember to apply Heisenberg's uncertainty principle formula: $\Delta x \Delta p \geq \frac{h}{4\pi}$ for position and momentum uncertainties.

Answer 1

The Correct Option is C

Given uncertainty in position, $\Delta x = 100$ pm = $100 \times 10^{-12}$ m = $10^{-10}$ m.
Planck's constant, $h = 6.626 \times 10^{-34}$ Js.
Using Heisenberg's uncertainty principle:
$\Delta x \Delta p \geq \frac{h}{4\pi}$, where $\Delta p$ is uncertainty in momentum.
Therefore, $\Delta p \geq \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 10^{-10}} = \frac{6.626 \times 10^{-34}}{1.25664 \times 10^{-9}} = 5.27 \times 10^{-25}$ kg m/s.
Rearranging to the form given in options, $\Delta p = 0.527 \times 10^{-24}$ kg m/s.