Question

Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 × 10⁵ ms⁻¹.

Answer 1

Heisenberg's Uncertainty Principle states that There is a fundamental and inherent limit of precision by which two quantites can be measured.

There is one equation, which states that The product of uncertainties in position and momentum cannot be less than a certain value.


This means that the more precisely we can measure one quantity, the less precisely we will know the values of other quantity. 

The mathematical form is:

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


Here, 

$\Delta x = \text{Uncertainty in Position} \\ \\ \Delta p = \text{Uncertainty in Momentum} \\ \\ h = \text{Planck's Constant}$

We can write momentum as 

$p = m v$

Uncertainty in momentum can also be written as:
$\Delta p = m \Delta v$


So we can also write our Uncertainty Equation as:

$\Delta x . (m \Delta v) \geq = \frac{h}{4\pi} \\ \\ \\ \implies \boxed{\Delta x. \Delta v = \frac{h}{4\pi m}}$

Now, here we have the following data from question:

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

The concerned particle is an electron. So we know its mass.

Our other data is:

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

For calculation purposes, we usually consider the minimum product of uncertainties, and we replace the $\geq$ sign with an equality one.


Now, we can find uncertainty in position:

$\Delta x . \Delta v = \frac{h}{4\pi m} \\ \\ \\ \implies \Delta x = \frac{h}{4 \pi m \Delta v} \\ \\ \\ \implies \Delta x = \frac{6.626 \times 10^{-34}}{4 \times \pi \times 9.1 \times 10^{-31} \times 5.7 \times 10^5} \\ \\ \\ \implies \Delta v \approx 1.016 \times 10^{-10} \, \, m \\ \\ \\ \implies \boxed{\Delta v \approx 1.02 \times 10^{-10} \, \, m}$


Thus, uncertainty in position is $1.02 \times 10^{-10} \, \, metres$


Hope it helps
Purva
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