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

`5.2728* 10^(-25)\ kgm/s`

Step-by-step explanation:

h = Planck's constant =
`6.626* 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=(h)/(4\pi)\\\Rightarrow \Delta p=(h)/(4\pi\Delta x)\\\Rightarrow \Delta p=(6.626* 10^(-34))/(4\pi* 10^(-10))\\\Rightarrow \Delta p=5.2728* 10^(-25)\ kgm/s`

The minimum uncertainty in its momentum is
`5.2728* 10^(-25)\ kgm/s`