Casimir Effect: Overview, Questions, Preparation

In 1927, a German physicist named Werner Heisenberg described the dual behaviour of matter and radiation. According to his statement, it is impossible to simultaneously determine the exact position and momentum(or velocity) of an electron.

Mathematically, it is written as

ΔxΔpx ≥ h/4π

or ΔxΔ(mvx) ≥ h/4π

or ΔxΔvx ≥ h/4πm

Here,  Δx = uncertainty in the position

        Δpx(or Δvx) = uncertainty in momentum(or velocity) of the particle.

         h = Planck's constant = 6.626x10-34 Js

To measure an electron, we illuminate it with light or electromagnetic radiation having a wavelength smaller than the dimension of the electron. The high momentum of the light photons(p=h/λ) would change the energy of the electrons. Even though we can find the electron position through this process yet its velocity after collision remains unknown.  

The trajectory of any object is determined by its position and velocity at any instant. If we know the velocity and position of a particle at a particular time, we can calculate its position and velocity after some time. Heisenberg's Uncertainty principle rules out the existence of such definite paths or trajectories of electrons and similar particles. The effect of this principle holds only for microscopic objects and is negligible for macroscopic objects.

Let us understand with the example of an electron.

Mass of electron = 9x10-31 kg

h = 6.626x10-34 Js

According to formula,

ΔxΔv = h/4πm

ΔxΔv = 6.626x10-34 Js/(4x3.1416x9x10-31)

ΔxΔv = 10-4m2s-1

It means that if the uncertainty in the position(Δx) of the electron is 10-8m, then the uncertainty in velocity(Δv) will be

10-4m2s-1/10-8m ≈ 104ms-1

This value is so large that the electrons moving in Bohr's orbit cannot hold good. Therefore, the precise statement about the position and momentum of electrons is replaced by the statement of probability. It is the main idea behind the quantum mechanical model of the atom.

The topic comes under the class XI NCERT Chemistry chapter 'Structure of Atom'. Along with the chapter 'Some basic concepts of Chemistry', it appears for 11 marks in the exams.

Example 1: What can we say about the uncertainty principle applied to a particle of milligram mass?

Solution: Given, m = 10-6kg

Also, we know that h = 6.626x10-34Js

Thus, ΔxΔv = h/4πm = 6.626x10-34/(4x3.1416x10-6) = 10-28m2s-1.

Since this value is extremely small, we consider it to be insignificant. Thus the uncertainties are almost negligible because the mass of the object is significantly large.

Example 2: A microscope locates an electron in an atom within a distance of 0.1Å. What is the uncertainty in velocity?

Solution: ΔxΔv = h/4πm or Δv = h/4πmΔx

Thus, Δv = 6.626x10-34/(4x3.14x0.1x10-10x9.11x10-31) = 5.79x106ms-1

Example 3: Given m = 40g and v = 45m/s. If the speed was measured within 2% accuracy, find the uncertainty in the position.

Solution: Uncertainty in speed = 45x2/100 = 0.9m/s

Δx = h/4πmΔv = 6.626x10-34/(4x3.14x40x10-3x0.9) = 1.46x10-33m.

The value is so small that the uncertainty is almost negligible. Hence, for macroscopic objects, the uncertainty is meaningless.

Q: What is Heisenberg's Uncertainty Principle?

Q: Write the formula for Heisenberg's Uncertainty Principle.

Q: What are Δx, Δv, h and m in the formula for Heisenberg's Uncertainty Principle?

Q: What is the significance of Heisenberg's Uncertainty Principle?

Q: What is the importance of Heisenberg's Uncertainty Principle?