Question

Calculate the uncertainty in the position of an electron .If the uncertainty in the velocity is 4.5×10 raised power 5m/s

Answer 1

$\textsf{\underline {\Large {Heisenberg's Uncertainty Principle}}}$ :

$\textsf{\underline {\large {Solution}}}$ :

Given, Uncertainty in velocity of an electron,$\mathsf{\Delta{v}}$ = $\mathsf{4.5\:{\times{{10}^{5}}}m{s} ^{-1}}$

According to Heisenberg's Uncertainty Principle,

$\boxed{\mathsf{\Delta{x\:{\times{\Delta{p} \:{\geq{\dfrac{h} {4{\pi}}}}}}}}}$

Or

$\boxed{\mathsf{\Delta{x\:{\times{\:m{\Delta{v} \:{\geq{\dfrac{h} {4{\pi}}}}}}}}}}$

Uncertainty in position $\Rightarrow{\mathsf{\Delta{x}\:{\geq{\dfrac{h}{4{\pi{m{\Delta{v}}}} }}}}}$

Since, we know that,

⚫h ( Planck's Constant ) = $\mathsf{6.6\:{\times{{10}^{-34}}}Js}$

⚫Mass of an electron, m = $\mathsf{9.1\:{\times{{10}^{-31}}} \:kg}$

Now, Putting these values,

$\Rightarrow{\mathsf{\Delta{x}\:{\geq{\dfrac{h}{4{\pi{m{\Delta{v}}}} }}}}}$

$\Rightarrow{\mathsf{\Delta{x}\:=\:{\dfrac{6.6\:{\times{{10}^{-34}kg\:{m}^{2}\:{s}^{-1}}}}{4\:{\times{3.14\:{\times({9.1\:{\times{{10}^{-31}\:kg\:)\:{\times{(\:4.5\:{\times{{10}^{5}\:m\:{s}^{-1})}}}} }}}}}}}}}}}$

$\Rightarrow{\boxed{\mathsf{\Delta{x}\:=\: 1. 283\:{\times{{10}^{-10}}}m}}}$