Question

The speed of a certain electron is 995 km s−1 . If the uncertainty in its momentum is to be reduced to 0.0010 per cent, what uncertainty in its location must be tolerated?

Answer 1

Answer:

The uncertainty in the location that must be tolerated is $1.163 * 10^{-5} m$

Explanation:

From the uncertainty Principle,

Δ$_{y}$ Δ$_{p}$ $= \frac{h}{2\pi }$

The momentum P$_{y}$ = (mass of electron)(speed of electron)

                                = $(9.109 * 10^{-31}kg)(995 * 10^{3} m/s)$

                                = $9.0638 * 10^{-25}kgm/s$

If the uncertainty is reduced to a 0.0010%, then momentum

                              = $9.068 * 10^{-30}kgm/s$

Thus the uncertainty in the position would be:

                              Δ$_{y} = \frac{h}{2\pi } * \frac{1}{9.068 * 10^{-30} }$

                              Δ$_{y} \geq 1.163 * 10^{-5}m$