URGENT:

An electron has a velocity of 300 m s^-1 that is accurate to 0.01%. With what accuracy can one locate the position of this electron ( mass of electron = 9.1 * 10^-31 kg )?

We know,
momentum, p = m x v
 where m is the mass and v is the velocity of the particle.
Now,
$∆p = m∆v$
where, $∆p$ is uncertainty in momentum and $∆v$ is uncertainty in speed
Substituting the values for mass and uncertainty in speed of electron:
$$\begin{gathered} ∆p =9.1 x {10}^{-31}x \frac{0.01}{100}x300 \\ = 2.7 x {10}^{-32} Kg.m/s \end{gathered}$$
According Heisenberg's principle:
$$\begin{gathered} ∆\mathrm{x} x ∆p\ge \frac{h}{4\mathrm{\pi }} \\ here, ∆\mathrm{x} is the uncertainty in position \\ ∆x\ge \frac{h}{4\mathrm{\pi x}∆\mathrm{p}} \\ ∆x\ge \frac{6.626 x {10}^{-34}}{4\mathrm{\pi x}2.7 x {10}^{-32}} \\ ∆x\ge 1.95 x {10}^{-3} m or 1.95 mm \end{gathered}$$

Hence, the accuracy with which the position of the electron can be located is 1.95 mm