JEE Questions on Heisenberg’s Uncertainty Principle (with solution)
Question: The position of an electron is known to an accuracy of 10-10 m. What is the minimum uncertainty in its velocity?
Given h = 6.63×10-34 JS and me = 9.1×10-31 kg.
Solution:
We know Heisenberg’s Uncertainty Principle
Δx⋅Δp ≥ ℏ/2
Where:
- Δx = Uncertainty in position
- Δp = Uncertainty in momentum
- ℏ = Reduced Planck constant
Given Δx=10-10 m, ℏ=6.63×10-34, and find Δp=?.
Δp ≥ ℏ/2Δx
Δp ≥ 6.63×10-34/2×10-10 kg m/s
Δp≥3.315×10-24 kg m/s
This is the minimum uncertainty in momentum. Now, using the relation p=mv, we can find the minimum uncertainty in velocity. Since
9.1×10-31 kg for an electron.
3.315×10-24 = (9.1×10-31)v
v = 3.315×10-24/9.1×10-10 m/s
v ≈3.64×106 m/s
So, the minimum uncertainty in velocity of the electron is 3.64×106 m/s.
Heisenberg Uncertainty Principle – Definition, Equation, Significance
Heisenberg Uncertainty Principle is a basic theorem in quantum mechanics. It state that we can not measure position and momentum of a particle both at the same time with the same accuracy. It means that if we try to measure the accurate position of a particle, then at the same time we can’t accurately measure the momentum of the particle. Mathematically, the product of uncertainties in position and momentum is greater than h/4π, where h is Planck’s constant. The principle is named after Werner Heisenberg, who proposed this theory in 1927.
In this article, we will learn in detail about Heisenberg’s Uncertainty Principle, its origin, formula, derivation, and other equations related to it. We will also learn its importance, applications, and other related concepts.