Question

In an atom, an electron is confined to a space of roughly 10−10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty Δp in its momentum? Express your answer in kilogram meters per second to two significant figures.

Answer 1

Answer:

Δp = 3.31 × 10⁻²⁴ kg.m/s

Explanation:

Given:

Electron is confined to a space of,  Δx  = 10⁻¹⁰ m

According to the Heisenberg uncertainty principle, we have

$\Delta p=\frac{h}{2\Delta x}$

where, h is the plank's constant = 6.626 × 10⁻³⁴ J.s

Δp is the uncertainty in the momentum

on substituting the respective values, we get

$\Delta p=\frac{6.626\times10^{-34}}{2\times10^{-10}}$

or

Δp = 3.31 × 10⁻²⁴ kg.m/s