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Question: Uncertainty in position and momentum are equal. Uncertainty in velocity is: A.\(\sqrt {h/\pi } \) ...

Uncertainty in position and momentum are equal. Uncertainty in velocity is:
A.h/π\sqrt {h/\pi }
B.h/2π\sqrt {h/2\pi }
C.1/2mh/π1/2m\sqrt {h/\pi }
D.None of these

Explanation

Solution

To answer this question, you must be familiar with the Heisenberg uncertainty principle. The position and velocity of an electron in an atom cannot be accurately determined and there is bound to be some uncertainty in the values. This uncertainty was given by the Heisenberg uncertainty principle.
Formula used:
The mathematical expression for the Heisenberg’s uncertainty principle is:
Δx.Δph4π\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }} or Δx.mΔvh4π\Delta x.m\Delta v \geqslant \dfrac{h}{{4\pi }}
Where, Δx\Delta x is the uncertainty in the position of electrons along a certain axis.
Δp\Delta p is the uncertainty in momentum parallel to the axis in consideration
mm is the mass of the electron
Δv\Delta v is the uncertainty in the velocity of the electron along the axis in consideration
And hh is the Planck’s constant

Complete step by step solution:
The Heisenberg uncertainty principle states that the determination of the exact position and momentum of a microscopic particle namely electron simultaneously is not possible with any degree of precision. We know the mathematical representation of the principle as
\Rightarrow Δx.Δph4π\Delta x.\Delta p \geqslant \dfrac{h}{{4\pi }}.
For ease of solving in numerical problems, we can consider the expression for the uncertainty principle as
\Rightarrow Δx.Δp=h4π\Delta x.\Delta p = \dfrac{h}{{4\pi }}.
Since momentum is the product of the mass and velocity of a particle, the expression can also be written as
\Rightarrow Δx.mΔv=h4π\Delta x.m\Delta v = \dfrac{h}{{4\pi }}
In the question, we are given that the uncertainty in the position is equal to the uncertainty in the momentum, that is, Δx=Δp\Delta x = \Delta p or Δx=mΔv\Delta x = m\Delta v
So we can write the expression as
\Rightarrow mΔv.mΔv=h4πm\Delta v.m\Delta v = \dfrac{h}{{4\pi }}
(mΔv)2=h4π{\left( {m\Delta v} \right)^2} = \dfrac{h}{{4\pi }}
Further solving for Δv\Delta v, we get
\Rightarrow Δv=12mhπ\Delta v = \dfrac{1}{{2m}}\sqrt {\dfrac{h}{\pi }}

Thus, the correct option is B.

Note:
It should be noted that the uncertainty principle is applied to microscopic particles along a fixed axis, that is, the uncertainties in position and momentum should be along the same axis. In simpler terms, we can say that if we know the momentum of a particle along a certain axis precisely, then the position of the particle is completely uncertain along that axis.