Question

The mathematical expression which is true for the uncertainty principle is:

• A. $\left(\Delta x\right)\left(\Delta v\right)\ge \frac{h}{4\pi }$
• B. $\left(\Delta E\right)\left(\Delta x\right)\ge \frac{h}{4\pi }$
• C. $\left(\Delta \theta \right)\left(\Delta \varphi \right)\ge \frac{h}{4\pi }$
• D. $\left(\Delta x\right)\left(\Delta m\right)\ge \frac{h}{4\pi }$

Answer 1

Answer: (A)

$\left(\Delta x\right)\left(\Delta v\right)\ge \frac{h}{4\pi }$

Heisenberg uncertainty principle states that, it is not possible to determine simultaneously the position and momentum of a moving microscopic particle with absolute accuracy if $\Delta p$ is the uncertainty in the determination of the position and $\Delta p$ is the uncertainty in the determination of momentum of a very small moving particle, then according to Heisenberg,

$\Delta x\Delta p=\frac{h}{4\pi }$

Where is the Planck's constant the above equation can also be expressed as,

$$\Delta x\Delta v=\frac{h}{4\pi m}$ , because $\Delta p=m\Delta v$

where $\Delta v$ is the uncertainty in the determination of velocity and m is the mass of the body. From the equation it is clear that the minimum uncertainty in the simultaneous determination of position and velocity is equal to, $\frac{h}{4\pi m}$

And, smaller the mass of the particle greater is the uncertainty.

More ever, if $\Delta x$ is very small, that is the position of a particle is known more or less exactly, $\Delta v$ would be large, that is uncertainty in the determination of velocity would be large similarly, if an attempt is made to measure exactly the velocity of the particle uncertainty with respect to the position will be large.