Atomic Models

# Heisenberg Uncertainty Principle

A German physicist, Werner Heisenberg, proposed the Uncertainty Principle in 1927. This principle is about the uncertainties in simultaneous measurements of position and momentum of small particles. It is known as Heisenberg’s Uncertainty Principle. This principle is the direct consequence of wave particle duality.

In classical mechanics, a moving particle has a definite momentum and occupies a definite position in space. So, it is possible to determine both its position and velocity (or momentum).

In quantum mechanics, a moving particle is described by a wave group (wave packet). So, the particle may be located anywhere within the group at a given time. For a small wave group, the position of the particle can be fixed but the particle will spread rapidly and hence the velocity becomes uncertain. For a large wave group, the velocity can be fixed but there is large indefiniteness in position. In this way, the certainty in position involves uncertainty in velocity or momentum and vice versa.

Heisenberg Uncertainty Principle states that it is impossible to measure simultaneously the position and momentum of a small particle with absolute accuracy or certainty. If an attempt is made to measure any one of these two quantities with greater accuracy, the other becomes less accurate.

If $\Delta x$ is the uncertainty in the position and $\Delta p$ is the uncertainty in the momentum of a particle, then the uncertainty principle is mathematically defined as $\Delta x\cdot\Delta p ≥ \frac{h}{4π}\text{ __(1)}$ where $h$ is the Planck’s constant.

This is the mathematical statement of Heisenberg Uncertainty Principle. It is also called indeterminacy principle.

We have, $\Delta p=m\Delta v$, where $m$ is the mass of the particle and $\Delta v$ is the uncertainty in velocity of the particle.

$\therefore \Delta x\cdot(m\Delta v)≥\frac{h}{4π}$ $\Delta x\cdot\Delta v≥\frac{h}{4πm}\text{ __(2)}$

Equation $(1)$ can also be written as, $\Delta x\cdot\Delta p≥\frac{\hbar}{2}\text{ __(3)}$

where, $\hbar=\frac{h}{2π}$. This mathematical relation is applicable to all those physical quantities whose pair product gives the dimension of Planck’s constant.

So, it can also be written in terms of energy and time as $\Delta E\cdot\Delta t≥\frac{\hbar}{2}$ Also, in terms of number and phase $\Delta n\cdot\Delta\phi ≥\frac{\hbar}{2}$

## Significance of Uncertainty Principle

Suppose we attempt to measure both the position and momentum of an electron. To locate the position of the electron, we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. As a result of the hitting by the photon, the position as well as the velocity of the electron are altered.

It should be noted that the uncertainty is not due to lack of sufficiently defined techniques. It is due to the fact that we cannot observe microscopic things without disturbing them. No instrument can observe the position of an electron without affecting its motion. Uncertainty principle is the fundamental limitation of nature.

If an attempt is made to determine the position of a moving electron by throwing light on it, its position cannot be determined with accuracy better than $\pm\lambda,$ the wavelength of light. If the light is of very small wavelength than the momentum $(h/\lambda)$ of the corresponding photon becomes so large that it would change the velocity of the moving electron thus introducing larger uncertainty in its position.

Although Heisenberg’s Uncertainty Principle holds good for all objects but it has no significance in our daily life. This is because, in our daily life, we interact with larger objects which we can observe with our naked eyes without altering their motion. Obviously the energy of the photon is insufficient to change the position and velocity of larger objects when it collides with them. For example, the light from a torch falling on a running ball in a dark room neither changes the speed of the ball nor its direction i.e. position.

The value of Planck’s constant is so small that the limitations imposed by the uncertainty principle are significant only for microscopic particles. Let us illustrate this further by taking examples of a mass of $1\text{ g}$ and an electron. For a mass of $1\text{ g},$ the product $\Delta x\cdot\Delta v$ comes extremely small. Hence, the uncertainties are negligible. However for an electron of mass $9.1×10^{-31}\text{ kg},$ the value of $\Delta x\cdot\Delta v$ is quite insignificant.

## Applications of Uncertainty Principle

One of the important implications of the Heisenberg’s Uncertainty Principle is that it rules out the well defined circular orbits proposed by Bohr’s atomic model. Since for a subatomic particle like an electron, it is not possible to determine its position and momentum simultaneously with good accuracy. Therefore, it is not possible to talk about the well defined circular orbits.

The Uncertainty Principle explains a large number of facts which could not be explained by classical ideas. It also explains that an electron cannot exist inside a nucleus. The radius of nucleus of any atom is of the order of $10^{-14}\text{ m}$ so that if an electron is confined within nucleus, the uncertainty in its position must not be greater than $10^{-14}\text{ m}.$

$\therefore\Delta x\cdot\Delta p≥\frac{\hbar}{2}$ $\Delta p≥\frac{\hbar}{2\Delta x}$ $\Delta p≥\frac{1.054×10^{-34}}{2×10^{-14}}$ $\Delta p≥5.27×10^{-21}\text{ kgms}^{-1}$

If this is the uncertainty in the nuclear electron’s momentum, the momentum $p$ itself must be at least comparable with this magnitude i.e. $p=5.27×10^{-21}\text{ kgms}^{-1}$

An electron with such a momentum has a kinetic energy many times greater than its rest mass energy $m_0c^2.$

The kinetic energy of electron of mass $m$ is given by $K.E.=\frac{p^2}{2m}=\frac{(5.27×10^{-21})^2}{2×9.1×10^{-31}}\text{ Joule}$ $=\frac{(5.27×10^{-21})^2}{2×9.1×10^{-31}×1.6×10^{-19}}\text{ eV}$ $=9.7×10^7\text{ eV}=97\text{ MeV}$

This means that if the electrons exist inside the nucleus, their K.E. must be of the order of $97\text{ MeV}$. But experimental observations show that no electron in the atom possesses energy greater than $4\text{ MeV}$. Therefore, it is confirm that the electrons do not exist inside the nucleus.