## HEISENBERG’S UNCERTAINTY PRINCIPLE

All moving objects that we see around us e.g., a car, a ball thrown in the air etc., move along definite paths. Hence their position and velocity can be measured accurately at any instant of time. Is it possible for subatomic particle also?

As a consequence of dual nature of matter, Heisenberg, in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass x velocity) of small particles.

This Principle States:“It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty” i.e., if an attempt is made to measure any one of these two quantities with higher accuracy, the other becomes less accurate.

The product of the uncertainty in position (x) and the uncertainty in the momentum (p = m.v where m is the mass of the particle and v is the uncertainty in velocity) is equal to or greater than h/4π where h is the Planck’s constant.

Thus, the mathematical expression for the Heisenberg’s uncertainty principle is simply written as

x.p > h/4π

Explanation of Heisenberg’s uncertainty principleSuppose we attempt to measure both the position and momentum of an electron, to pinpoint 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, the position as well as the velocity of the electron are disturbed. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used. The uncertainty in position is ±λ. The shorter the wavelength, the greater is the accuracy. But shorter wavelength means higher frequency and hence higher energy. This high energy photon on striking the electron changes its speed as well as direction. But this is not true for macroscopic moving particle. Hence Heisenberg’s uncertainty principle is not applicable to macroscopic particles.

Illustration:Why electron cannot exist inside the nucleus according to Heisenberg’s uncertainty principle?

Solution:Diameter of the atomic nucleus is of the order of 10

^{–15}mThe maximum uncertainty in the position of electron is 10

^{–15}m.Mass of electron = 9.1 x 10

^{–31}kg.x. p = h / 4π

x x (m.v) = h/4π

v = h / 4π x 1 / x.m = 6.63 x 10

^{-34}/ 4 x (22/7) x 1 / 10^{-15}x 9.1 x 10^{-31}v = 5.80 x 10

^{10}ms^{–1}This value is much higher than the velocity of light and hence not possible.

Illustration:What is the uncertainty in the position of electron, if uncertainty in its velocity is 0.0058 m/s?

Solution:x x v = h / 4πm

x = 6.02 x 10

^{-34}/ 4 x 3.14 x 9.1 x 10^{-34}x 0.0058x = 0.01 m