Question

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

Answer 1

Hint: Heisenberg’s uncertainty principle states that the position and momentum of a particle cannot be determined simultaneously with precision. Mathematically, it is expressed as $\mathrm{\Delta }x\cdot \mathrm{\Delta }{p}_x⩾{\displaystyle \frac{h}{4\pi }}$. Here, $\mathrm{\Delta }x$ is the uncertainty in position and $\mathrm{\Delta }{p}_x$ is the uncertainty in momentum.

Complete step by step solution:
We know that the Heisenberg’s uncertainty principle is mathematically expressed as,
$\mathrm{\Delta }x\cdot \mathrm{\Delta }{p}_x⩾{\displaystyle \frac{h}{4\pi }}$ or $\mathrm{\Delta }x\cdot \mathrm{\Delta }p={\displaystyle \frac{h}{4\pi }}$
where $\mathrm{\Delta }x$ is the uncertainty in position,
             $\mathrm{\Delta }{p}_x$ is the uncertainty in momentum,
             $h$ is the Planck’s constant.
We are given that the uncertainty in position and momentum are equal. Thus,
$\mathrm{\Delta }x=\mathrm{\Delta }{p}_x$
We know that the momentum of a particle is the product of the mass of the particle and the velocity with which the particle is moving. Thus,
$\mathrm{\Delta }p=m\mathrm{\Delta }v$
where $\mathrm{\Delta }{p}_x$ is the uncertainty in momentum,
             $m$ is the mass of the particle,
             $\mathrm{\Delta }v$ is the velocity of the particle.
Thus, the mathematical expression for Heisenberg’s uncertainty principle is,
$\mathrm{\Delta }x\cdot m\mathrm{\Delta }v={\displaystyle \frac{h}{4\pi }}$
We are given that the uncertainty in position and momentum are equal. Thus,
$\mathrm{\Delta }x=m\mathrm{\Delta }v$
Thus, the mathematical expression for Heisenberg’s uncertainty principle is,
$m\mathrm{\Delta }v\cdot m\mathrm{\Delta }v={\displaystyle \frac{h}{4\pi }}$
$\mathrm{\Delta }{v}^2={\displaystyle \frac{h}{m^{2}4\pi }}$
Take the square root on both sides of the equation. Thus,
$\mathrm{\Delta }{v}^2={\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$
Thus, the uncertainty in velocity is ${\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$.
Thus, if uncertainty in position and momentum are equal then uncertainty in velocity is ${\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$.

Thus, the correct option is (C) ${\displaystyle \frac{1}{2m}}\sqrt{{\displaystyle \frac{h}{\pi }}}$.

Note:
Heisenberg’s uncertainty principle is not applicable to the macroscopic particles but it is applicable only to the microscopic particles i.e. the uncertainty in position and the uncertainty in momentum must be along the same axis. We can say that if the momentum is parallel to an axis is precisely known then the position along the same axis is uncertain or vice versa.