Answer:
18931.4
Explanation:
Given : velocity of the electron = 2.0 [tex]\times[/tex]10
mass of the electron = 9.19[tex]\times[/tex] 103
we know that reduced planks constant, h = 6.5821[tex]\times[/tex] [tex]10^{-16}[/tex] eV s
We know from uncertainity principle,
[tex]\Delta \textup{x}.\Delta \textup{v} = \frac{h}{\dot{m}}[/tex]
[tex]\Delta \textup{x} = \frac{h}{\dot{m}\times \Delta \textup{v}}[/tex][tex]\Delta \textup{x} = \frac{6.5821\times 10^{-16}}{9.19\times 103\times 2.0\times 10}[/tex]
[tex]\Delta \textup{x}[/tex] = 18931.4 m
Hence, uncertainty in position of the electron is 18931.4
