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# Uncertainty Principle Formula

Quantum mechanics is the discipline of science, which is dealing with the measurements on the minuscule scale. That measurements are useful in macro and microphysics and can lead to very diverse consequences. Heisenberg uncertainty principle or simply uncertainty principle is a very important concept in Quantum mechanics. This article will discuss the uncertainty principle as well Uncertainty Principle formula with examples. Let us begin it!

## Heisenberg Uncertainty Principle

This principle was given in 1927 by the German physicist Werner Heisenberg. This principle says that the position and momentum of any particle cannot be simultaneously measured with arbitrarily high accuracy. And, there exists a minimum value for the product of the uncertainties of these two measurements. Hence there is also a minimum for the product of the uncertainties of energy and time. It is due to the wave properties inherent in the quantum mechanical description of nature.

Source: en.wikipedia.org

The uncertainty is much inherent property of nature. Thus we may conclude that we cannot determine both the position and momentum of a particle at the same time and accurately. We may also conclude that the very concepts of exact position and exact velocity together, in fact, have no meaning in nature.

Ordinary experience in science will not provide any clue on this principle. It is due to the fact that it is easy to measure both the position and the velocity of any object. This is because that the uncertainties implied by this principle for ordinary objects are too small to be observed.

Therefore, the product of the uncertainties in position and velocity is equal to or greater than a very very small physical quantity i.e. h. So, only for the exceedingly small masses of atoms and subatomic particles thus product of the uncertainties will be significant.

### The Formula for Uncertainty Principle:

This principles’ statement is saying that,

$$(Position uncertainty) \times (momentum uncertainty) \geq \frac{(Planck’s constant)}{2}$$

And,

$$(Energy uncertainty) \times (time uncertainty) \geq \frac{(Planck’s constant)}{2}$$

The result of position and momentum will always greater than $$\frac{h}{4 \pi}$$. The formula for Heisenberg Uncertainty principle is given mathematically as:

$$\Delta x \times \Delta p \geq \frac{h}{4 \pi}$$

Where,

 h Planck’s constant , having value 6.626 \times 10^{-34} Js $$\Delta p$$ is the uncertainty in momentum $$\Delta x$$ is the uncertainty in position

## Solved Examples for Uncertainty Principle Formula

Q.1: The uncertainty in the momentum of a ball travelling at speed of $$20 m s^{-1} is 1 \times 10^{−6}$$ of its momentum. Calculate the uncertainty in position? Mass of the ball is given as 0.5 kg.

Solution: Known parameters in the problem are as follows,

Velocity,$$v = 20 m s^{-1}$$

Mass, m = 0.5 kg,

Plank’s Constant, h has a value of $$6.626 \times 10^{−34} J s$$

And, uncertainty in momentum is,

$$\Delta p = p \times 1 \times 10^{−6}, for momentum p.$$

As we know,

Momentum,$$p = m \times v$$

So, $$p = 0.5 \times 20$$

i.e. $$p = 10 Kg m s^{-1}$$

Now, $$\Delta p = p \times 1 \times 10^{−6}$$

i.e. $$\Delta p = 10 \times 1 \times 10^{−6}.$$

$$\Delta p = 1 \times 10^{-5}$$

Now, Heisenberg Uncertainty principle formula is:

$$\Delta x \times \Delta p \geq \frac{h}{4 \pi}\\$$

i.e., $$\Delta x \geq \frac{h}{4 \pi \times \Delta p}\\$$

$$\Delta x \geq \frac{6.626 \times 10^{-34}}{4\times 3.14 \times 10^{-5}}\\$$

$$\Delta x \geq 0.527 \times 10 ^{-29} m$$

Therefore, uncertainty in position will be$$0.527 \times 10 ^{-29} m.$$

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