Question

What is the uncertainty in position of an electron of an atom if there is t 2.0 x 10' msec uncertainty in its velocity? Use the reduced Planck's constant and electron mass 9.19 x 103 kg.

Answer 1

Answer:

18931.4

Explanation:

Given : velocity of the electron = 2.0 $\times$10

            mass of the electron = 9.19$\times$ 103

we know that reduced planks constant, h = 6.5821$\times$ $10^{-16}$ eV s

We know from uncertainity principle,

$\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}$

$\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}$$\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}$

$\Delta \textup{x}$ = 18931.4 m

Hence, uncertainty in position of the electron is 18931.4