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# Heisenberg Uncertainty Principle

Neils Bohr purposed the concept of atomic orbital and tried to overcome the limitations of the Rutherford atomic model. Bohr’s model put forward the some postulates related to the model of an atom. He used the concept of the emission spectrum of hydrogen.

In an atom, electrons can be assuming only certain orbits around the nucleus. These orbits are stable and known as “stationary" orbits or energy levels. When electrons revolve in these discrete orbits, they do not radiate any energy. Each orbit has a certain level of energy.  For example, the orbit closest to the nucleus has energy E1, the next closest E2 and so on. With these conditions Bohr was able to explain the stability of atoms. These energy levels also known as shells or ‘orbits‘ can be represented in two different ways: either by using numbers for them like ; 1, 2, 3, 4, 5 and 6 or by certain alphabets like; K, L, M, N, O and P. They are counted from centre to outwards, means from nucleus to electrons. As we move away from the nucleus, the energy of levels increase, hence nearest energy level has lowest energy.

Bohr's model successfully explained the limitation of Rutherford’s model and received world-wide fame. The main drawback of Bohr’s model was a limitation of his explanation of the hydrogen atom.  As he worked on hydrogen atom, couldn’t explain the energy levels in other atoms. Thus the final atomic model was yet to be developed.

## Heisenberg Uncertainty Principle Definition

Due to the consequences of dual nature of matter, German scientist “Werner Heisenberg” gave a principle about the uncertainties in simultaneous measurement of position and momentum of small particles. This is known as Heisenberg’s uncertainty principle.
The statement of the principle states that "It is impossible to measure the position and momentum of a small particle with absolute accuracy. If any attempt is doing to measure any one of these quantities with good accuracy then the other becomes less accurate."

## Heisenberg Uncertainty Principle Equation

The product of the uncertainty in the position (Δx) and the uncertainty in momentum (Δp) is always constant. This is equal to or greater than h/4p, where h is plank’s constant.
Δp = m

Δv where m is the mass of the particle and Δv shows the uncertainty in velocity.
Thus, the uncertainty equation is

Δp. Δx ≥ h/4p

## Heisenberg Uncertainty Principle Formula

The mathematical expression for the Heisenberg's uncertainty principle
$\Delta$ p. $\Delta$ x $\propto$ h/4p
By using the above equation, the calculated value of $\Delta$ p is the minimum value of $\Delta$ p for any particular value of $\Delta$ x. Similar the calculated value of $\Delta$ x also shows the minimum value of $\Delta$ x for any specific value of $\Delta$ p.
As $\Delta$p = m. $\Delta$ v,
So the equation becomes (m. $\Delta$ v). $\Delta$ x $\pi$ h/4p
Or
($\Delta$ v)($\Delta$ x) $\propto$ h/4pm

This equation shows that the position and velocity of particle cannot be measured simultaneously with accuracy.
• The uncertainty principle also applies for the same axis of position and momentum.
• Like if $\Delta$ x is along the y axis then $\Delta$ p also must be along the same y axis.
• The uncertainty principle also does not have any limitation for measuring instrument.

### Explanation of Uncertainty Principle

The main principle of Heisenberg uncertainty principle can be understood by following description.

1. For measuring the position and momentum of electron, a light phenomenon is used.
2. To know the position of electron, a photon of light is become strike on the electron.
3. The microscope is used to see the reflected photon.
4. Due to hitting with photon, both the position and momentum of electron is disturbed.
5. The electron has wave particle duality. It shows an intrinsic uncertainty position and momentum.
6. Due to small and light weight of electron, the incident photon changes the motion of electron after interacting with the electron.

But the principle of optics states that the accurate measurement for the position of particle depends on the wave length of used light. The uncertainty of position is equal to ± $\Delta$. The accuracy is increased with decreasing the wave length. But the small value of wave length shows the high frequency and thus the high energy. Thus the high energy photon alters the position and speed of electron. It can also be explained in terms of momentum.

The shorter wave length also shows the high value of momentum (p = h/ $\Delta$) and so the photon of high momentum transfers the indefinite and large amount of momentum to the electron. This produces the high uncertainty in the velocity or speed of electrons. Or we can also say that the wave length increases with decreasing in the value of momentum. This shows the uncertainty in position of electron.

The momentum of electron changes with interacting photon and if the initial motion of photon is unknown in the first place, the following two cases are possible.
1. There is no transfer of momentum between photon and electron.
2. The photon transfers all of its momentum to the electron.
After the collision, there is uncertainty in the momentum and position of the electron. This must be equal to the momentum of the photon (p). The position of electron is also not fixed.

## Significance of Heisenberg Uncertainty Principle

Heisenberg’s uncertainty principle is good for all the objects but is more significant for microscopic particles. The main reason behind this is that the energy of photon is enough to alter the position and momentum of electron or other microscopic particles.

This is not sufficient for bigger bodies. For example, the light on the running rat from the torch cannot change the direction and position of rat.
Or if mass of particle is 1 mg then by using Heisenberg’s uncertainty principle formula.

(Δv)(Δx) = (h/4) π m

By putting the value of h = 6.626 x 10-34 kg m2 s-1,
m = 1 mg or 10-6kg.
So (Δv) (Δx) = (6.626 x 10-34 / 4) x 3.1416 x 10-6,
Or (Δv) (Δx) = 10-28 m2 s-1

Thus the product of (Δv) and (Δx) is very small. The particles having mass greater than 1mg, the product will still be small. Hence, the values are negligible. But for the electron, the product is equal to (Δv) (Δx) = 10-4 m2 s-1.
It shows that if the uncertainty in position is 10-4m then the uncertainty in velocity is equal to 1 m2s. This value is measurable not negligible. This proves that the electrons have the probability of having a particular position and momentum. This is the basic of wave mechanical model of atom.

The Bohr concept of fixed path with definite position and momentum of electron has replaced by this fact. The Heisenberg’s uncertainty principle has not any kind of significance in daily life as in daily life the position and velocity can be measured with accuracy. This principle also proved about the non existence of electrons in the atomic nucleus. This is because the diameter of atomic nucleus is of the order of approx 10-15m and so if the electron would be existing in the atomic nucleus then its maximum uncertainty in position (Δx) should be equal to 10-15m.

The mass of electron is 9.1 x 10-31kg.

By putting these values in the formula of uncertainty, the uncertainty in velocity (Δv) is equal to

(Δv)(Δx) = (h/4) π m
Or (Δv) = (h/4) π m (Δx),
Or (Δv) = (6.626 x 10-34 / 4) x 3.1416 x 10-15 x 9.1 x 10-31
Or (Δv) = 5.77 x 1010 m s-1

This value of uncertainty in velocity is comparatively higher than the velocity of light (3 x 108 m s-1) which is not possible and thus, it proves that the electrons are not present in atomic nucleus.

## Heisenberg Uncertainty Principle Example

Below you could see example

### Solved Examples

Question 1: If the mass of an electron is 10-27kg and the uncertainty in position is equal to 10-11m then find the uncertainty in velocity.
Solution:

(Δv) (Δx) = ($\frac{h}{4}$) πm,

Put the value of Δx =10-11m,
h = 6.626 x 10-34 kg m2 s-1 and m =10-27kg.
So (Δv) = ($\frac{h}{4}$) π m (Δx),
Or
Δv = $\frac{6.626 x 10^{-34}}{ 4 \times 3.1416 \times 10^{-27} \times 10^{-11}}$

Δv = 5.27 x 105 m s-1

Question 2: Find out the accuracy in position of a particle if the uncertainty in the momentum of the particle is 2.2 x 10-4g cm s-1. (h= 6.626 x 10-27 ergs)
Solution:

(Δp) (Δx) = ($\frac{h}{4}$) π

So (Δx) = $\frac{6.626 x 10^{-27}}{ 4 \times 3.1416 \times 2.2 \times10^{-4}}$

Or
Δx = 2.39 x 10-24cm

Question 3: Calculate the uncertainty in position if the electron has a speed of 500 m s-1 with an uncertainty of 0.02%.
Solution:

The uncertainty in speed (Δv) = ($\frac{0.02}{100}$) x 500 m s-1 = 0.1 m s-1

Now by formula, -(Δv) (Δx) = ($\frac{h}{4}$) π  m

Or (Δx) = ($\frac{h}{4}$) π  m (Δv), h = 6.626 x 10-34 J s, m = 9.1 x10-31 kg, Δv = 0.1 m s-1`

Or Δx = ($\frac{6.626 \times 10^{-34}}{4}$) x 3.1416 x 9.1 x10-31 x 10-1
So
Δx = 5.78 x 10-4m

Question 4: A particle has mass of 3000kg with the accuracy in position of ±10 pm. Find out its uncertainty in velocity.
Solution:

m = 3000kg, Δx = 10pm = 10 x 10-10; m  = 10-11m.
By uncertainty principle (Δv) = ($\frac{h}{4}$) πm (Δx),

So Δv = ($\frac{6.626 \times 10^{-34}}{4}$) x 3.1416 x 3000 x10-11
Or
Δv = 1.75 x 10-27 m s-1

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