Question

If the uncertainties in position and momentum are equal, then uncertainty in velocity is:
(A) ${\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$
(B) $\sqrt{{\displaystyle \frac{h}{2\pi }}}$
(C) ${\displaystyle \frac{1}{m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$
(D) $\sqrt{{\displaystyle \frac{h}{\pi }}}$

Answer 1

Hint: Heisenberg’s uncertainty principle states that it is not possible to determine the position as well as momentum of a microscopic particle simultaneously with any degree of precision. Mathematically, it is given as
     $\mathrm{\Delta }x.\mathrm{\Delta }{p}_x\ge {\displaystyle \frac{h}{4\pi }}$
Where $\mathrm{\Delta }x$ and $\mathrm{\Delta }{p}_x$ are the uncertainties in the position and momentum, respectively.

Complete step by step answer:
We know that the mathematical expression for the for the Heisenberg’s principle is given as
     $\mathrm{\Delta }x.\mathrm{\Delta }{p}_x\ge {\displaystyle \frac{h}{4\pi }}$
$\mathrm{\Delta }x$ is the uncertainty in position along an axis, say x-axis
$\mathrm{\Delta }{p}_x$ is the uncertainty in momentum parallel to the x-axis
$h$ is Planck’s constant and it value is $6.626\times {10}^{-34}Js$
We have been given that $$\mathrm{\Delta }x$$ and $\mathrm{\Delta }p$ are equal, i.e. $\mathrm{\Delta }x=\mathrm{\Delta }p$.

For our convenience, we can write the expression for Heisenberg’s uncertainty in numerical problems as
     $\mathrm{\Delta }x.\mathrm{\Delta }p={\displaystyle \frac{h}{4\pi }}$
Now we know that momentum, p is the product of the mass of the particle, m and the velocity, v with which the particle is moving.

Thus, we can write $\mathrm{\Delta }p=m\mathrm{\Delta }v$.
Using $\mathrm{\Delta }p=m\mathrm{\Delta }v$, now we have
     $\begin{array}{rl}& \mathrm{\Delta }x.\mathrm{\Delta }p={\displaystyle \frac{h}{4\pi }}\\ & \mathrm{\Delta }x.m\mathrm{\Delta }v={\displaystyle \frac{h}{4\pi }}\end{array}$

But it is given that $\mathrm{\Delta }x=m\mathrm{\Delta }v$ , therefore, the above equation becomes
     $\begin{array}{rl}& m\mathrm{\Delta }v.m\mathrm{\Delta }v={\displaystyle \frac{h}{4\pi }}\\ & {(m\mathrm{\Delta }v)}^2={\displaystyle \frac{h}{4\pi }}\end{array}$

Simplifying the above equation, we obtain
     $$\begin{array}{rl}& m^2\mathrm{\Delta }{v}^2={\displaystyle \frac{h}{4\pi }}\\ & \mathrm{\Delta }{v}^2={\displaystyle \frac{h}{m^{2}4\pi }}\end{array}$$

Now, taking square root on both sides, we get $$\mathrm{\Delta }v$$ as
     $$\begin{array}{rl}& \sqrt{\mathrm{\Delta }{v}^2}=\sqrt{{\displaystyle \frac{h}{m^{2}4\pi }}}\\ & \sqrt{\mathrm{\Delta }{v}^2}=\sqrt{{\displaystyle \frac{h}{m^2{2}^2\pi }}}\\ & \mathrm{\Delta }v={\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}\end{array}$$
Therefore, the uncertainty in the velocity is measured to be $$\mathrm{\Delta }v={\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$$. So, the correct answer is “Option A”.

Note: Carefully solve the question and do not make any errors. It is to be noted here that uncertainty principle is applied to microscopic particles along the same axis, i.e. $\mathrm{\Delta }x$ and $\mathrm{\Delta }p$ have along the same axis. In other words, we can say that if the momentum (or velocity) parallel to an axis is known precisely, then the position of the particle along that axis is uncertain completely.