Answer
2.8 * $10^{3}$ m$s^{-1}$, 0.56%
Work Step by Step
We have the formula according to Heisenberg uncertainty principle :
($\Delta$x).($\Delta$p) >= $\frac{h}{4π}$
($\Delta$p) = m.($\Delta$v)
First we need to get ($\Delta$p) so we can get the uncertainty in the velocity of an electron
($\Delta$p) >= $\frac{h}{4π \Delta x}$
($\Delta$p) = $\frac{6.626 * 10^{-34} Js}{4 *3.14 * 2 * 10^{-8}m}$
($\Delta$p) = 2.6 * $10^{-27}$ Js$m^{-1}$
And now we can get the uncertainty in the velocity of an electron which has a mass of 9.11 * $10^{-31}$ kg
($\Delta$p) = m.($\Delta$v)
($\Delta$v) = $\frac{\Delta p}{y}$
($\Delta$v) = $\frac{2*6*10^{-27}Jsm^{-1}}{9*11*10^{-31}kg}$
($\Delta$v) =2.8 * $10^{3}$ m$s^{-1}$
If electron is moving at a speed of 5.0 * $10^{5}$ m$s^{-1}$, the fraction of the speed is $\frac{2.8. 10^{3} ms^{-1}}{5.0 * 10^{5} ms^{-1}}$
= 5.6 * $10^{-3}$ = 0.56%
