Question

An electron has an uncertainty in its position of 587 pm . part a what is the uncertainty in its velocity?

Answer 1

We can solve the problem by using Heisenberg's uncertainty principle, which states that:
$\Delta x \Delta p \geq \frac{h}{4 \pi}$
where
$\Delta x$ is the uncertainty on the position
$\Delta p$ is the uncertainty on the momentum
$h$ is the Planck constant.

Keeping in mind that the momentum is the product between the mass of the electron and its velocity: 
$p=mv$
we can rewrite the Heisenberg principle as
$m \Delta x \Delta v \geq \frac{h}{4 \pi}$
where $\Delta v$ is the uncertainty on the velocity.

The uncertainty on the position is $\Delta x = 587 pm = 587 \cdot 10^{-12} m$, so we can find the uncertainty on the velocity by re-arranging the previous equation:
$\Delta v \geq \frac{h}{4 \pi m \Delta x}= \frac{6.6 \cdot 10^{-34} Js}{4 \pi (9.1 \cdot 10^{-31}kg)(587 \cdot 10^{-12} m)} =9.84 \cdot 10^4 m/s$