Question

Calculate the uncertainty in position of an electron if the uncertainty in the velocity is 5.7*10^5 msec^-1

Answer 1

Heisenberg's Uncertainty Principle states that there is a fundamental limit to the precision with which some quantities can be measured.

In the Uncertainty Relation of Position and Momentum, the better you know one quantity, the lesser precision you have with the other. 


The product of position and momentum of a quantum particle is never less than a certain value. The mathematical form is:

$\displaystyle \Delta x . \Delta p \geqslant \frac{h}{4\pi}$

Here,

$\Delta x$ = Uncertainty in Position
$\Delta p$ = Uncertainty in Momentum


We can write Momentum as 

$p=mv$

And since the mass of an object is constant, we can write:

$\Delta p = m\Delta v$

And we can rewrite the Heisenberg Equation as:

$\displaystyle \Delta x . m\Delta v \geqslant \frac{h}{4\pi} \\ \\ \\ \implies \Delta x .\Delta v \geqslant \frac{h}{4\pi m}$
_______________________


Here, our data is:

$\Delta v = 5.7 \times 10^5 \, \, m/s$

Also, the particle under consideration is an electron. And we also know the value of Planck's Constant. Our additional data is:

$m = 9.1 \times 10^{-31} \, \, kg \\ \\ h = 6.626 \times 10^{-34} \, \, Js$


We can find uncertainty in position as follows:


$\displaystyle \Delta x . \Delta v \geqslant \frac{h}{4\pi m} \\ \\ \\ \implies \Delta x \geqslant \frac{h}{4\pi m\Delta v} \\ \\ \\ \implies \Delta x \geqslant \frac{6.626 \times 10^{-34}}{4\pi \times 9.1 \times 10^{-31}\times 5.7 \times 10^5} \\ \\ \\ \implies \boxed{\bold{\Delta x \geqslant 1.02 \times 10^{-7} \, \, m}}$


Thus, The Uncertainty in Position is at least $\bold{1.02 \times 10^{-7} \, \, m}$