Question

How do you calculate the uncertainty in velocity (in ${\text{m}}{\text{.}}{{\text{s}}^{ - 1}}$)of an electron (mass $9.11 \times {10^{ - 31}}\,{\text{kg}}$) under the conditions where the uncertainty in position is $4.782 \times {10^{ - 3}}\,{\text{m}}$?

Answer 1

Hint: We know that Heisenberg’s uncertainty principle states that it is not possible to measure simultaneously the position and momentum of a microscopic particle with absolute accuracy or certainty.

Complete step by step answer:
Let’s understand the mathematical expression of Heisenberg’s uncertainty principle. Mathematically, the product of uncertainty in position and uncertainty in momentum of a microscopic particle is always constant and is equal to or greater than $dfrac{h}{4\pi}$.
$\Delta x.\Delta p \geqslant \dfrac{h}{4\pi} $…… (1)
Where, $\Delta x$ is uncertainty in measuring exact position and $\Delta p$ is uncertainty in measuring exact momentum.
From equation (1), it is evident that If $\Delta x$ is very small, the position of microscopic particles can be measured accurately, But $\Delta p$ will be very large which means that momentum or velocity of the particle cannot be measured with accuracy.
We know that $p = mv$ , where p is momentum, m is mass and v is velocity
So, we replace $\Delta p$ by $m\Delta v$. So, equation (1) becomes,
$\Delta x.m\Delta v \geqslant \dfrac{h}{4\pi}$
Now, we have to have to rearrange the above equation to calculate $\Delta x$.
\[\Delta v \geqslant \dfrac{h}
{{4\pi m\Delta x}}\]…… (2)
The mass of electron is $9.11 \times {10^{ - 31}}\,{\text{kg}}$
, the value of Planck’s constant is $6.626 \times {10^{ - 34}}\,{{\text{m}}^{\text{2}}}\,{\text{kg/s}}$
, the value of $\Delta x$
Is $4.782 \times {10^{ - 3}}\,{\text{m}}$
Now, we have to put all the above values in equation (2).
\[ \Rightarrow \Delta v = \dfrac{{6.626 \times {{10}^{ - 34}}\,{{\text{m}}^{\text{2}}}\,{\text{kg/s}}}}
{{4 \times 3.14 \times 9.11 \times {{10}^{ - 31}}\,{\text{kg}} \times 4.782 \times {{10}^{ - 3}}\,{\text{m}}}} = \dfrac{{6.626}}
{{574.16}} = 0.012\,m/s\]

Therefore, the uncertainty of velocity of the electron is 0.012 m/s.

Note: It is to be noted that the uncertainty principle has no impact on our daily life. It applies to the moving microscopic particles (protons, neutrons and electrons) which we can not see with our eye. It has no influence on the moving macro or semi micro objects which we actually see or observe. Thus, the Heisenberg’s uncertainty principle has no impact on our daily life.