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 ?p in its momentum?

Answer 1

Answer:

$5.2728\times 10^{-25}\ kgm/s$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

$\Delta x$ = Uncertainity in position = $10^{-10}\ m$

$\Delta p$ = Uncertainty in momentum

According to the Heisenberg uncertainity principle we have

$\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s$

The minimum uncertainty in its momentum is $5.2728\times 10^{-25}\ kgm/s$