Question

Calculate the uncertainty of the velocity of an electron, if the position uncertainty is 2.52×10^-11 m.

Answer 1

$$\Delta\chi\Delta p_x\geqslant h$$

$$\Delta\chi m_{e^-}\Delta v = h$$

Where $m_{e^-} = 9.109*10^{-31}kg, h = 6.626*10^{-34}J*sec$

Then

$$\Delta v = {h \over \Delta\chi m_{e^-}} = {6.626*10^{-34}J*sec \over 2.52*10^{-11}m*9.109*10^{-31}kg} = 0.289*10^8 m/sec = 2.89*10^7m/sec$$

Answer: 2.89*10⁷ m/sec.