• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar

Class Notes

Free Education for All

  • Class 1-5
  • Class 6
  • Class 7
  • Class 8
  • Class 9
  • Class 10
  • Class 11
  • Class 12
  • NCERT Solutions
    • NCERT Books
You are here: Home / Class 11 / Chemistry / Structure of Atom / Heisenberg’s Uncertainty Principle

Heisenberg’s Uncertainty Principle

Last Updated on June 26, 2022 By Mrs Shilpi Nagpal 4 Comments

​Werner Heisenberg, a German physicist, in 1927 gave a principle about the uncertainty in simultaneous measurement of position and momentum of small particles.

Heisenberg’s uncertainty Principle states that:

It is impossible to measure simultaneously the position and momentum of a small particle with absolute accuracy or certainty. The product of the uncertainty in the position ( Δx ) and the uncertainty in the momentum(Δp) is always constant and is equal to or greater than h/4π, where h is the Planck’s Constant ie.

Heisenberg's Uncertainty Principle

Significance of Heisenberg Uncertainty Principle

Heisenberg’s uncertainty Principle holds good for all objects but it is significant only for microscopic particles. The energy of the photon is insufficient to change the position and velocity of bigger bodies when it collides with them.

For a particle of mass 1 mg, we have

Δx . Δv = h / 4π m

Δx . Δv = 6.626 × 10-34 / 4 × 3.1416 × ( 106 )

Δx . Δv = 10-28 m2 s-1

The product Δx  and Δv is extremely small. For particle of mass greater than 1 mg, the product will be still smaller. Hence these values are negligible.

For a microscopic particle like an electron, we have

Δx . Δv = 6.626 × 10-34 / 4 × 3.1416 × ( 9.11 × 10-31)

Δx . Δv = 10-4 m2 s-1

If uncertainty in position is 10-4 ,uncertainty in velocity will be 0.1 m/s.

Bohr’s concept of fixed circular path with definite position and momentum of electron have been replaced by stating that the electron has the probability of having a given position and momentum.

Electron cannot exist in the nucleus

This is because the diameter of the atomic nucleus is of the order 10-15 m. If the electron were to exist within the nucleus, the maximum uncertainty in its position would have been 10-15 m .Taking the mass of electron as  9.1 × 10 -31kg, the minimum uncertainty in velocity can be calculated by applying uncertainty Principle as follow:

Δ x . Δ p = h/4π

Δ x. (m × Δv) = h/4π

Δ v = h/ 4π × Δx × m

Δ v = 6.626 × 10-34 / 4 × 3.1416 × 10-15 × 9.1 × 10 -31

Δ v = 5.77 × 1010  m/s

This value is much higher than the velocity of light ( 3 × 108 )and hence is not possible.

Filed Under: Structure of Atom, Chemistry, Class 11 Tagged With: Electron cannot exist in the nucleus, Heisenberg's uncertainty Principle

About Mrs Shilpi Nagpal

Author of this website, Mrs Shilpi Nagpal is MSc (Hons, Chemistry) and BSc (Hons, Chemistry) from Delhi University, B.Ed (I. P. University) and has many years of experience in teaching. She has started this educational website with the mindset of spreading Free Education to everyone.

Reader Interactions

Comments

  1. Renu goyal says

    January 4, 2018 at 5:13 am

    It is very helpful.

    Reply
    • Mrs Shilpi Nagpal says

      January 4, 2018 at 10:08 am

      Thanks renu

      Reply
  2. Venomtech says

    February 19, 2019 at 11:00 am

    That’s really helpful

    Reply
  3. Shreya says

    September 5, 2021 at 1:27 pm

    Thanks alot mam

    Reply

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • Chapter 10 The Beggar
  • Chapter 8 A House is Not a Home
  • Chapter 7 The Last Leaf
  • Chapter 6 Weathering the Storm in Ersama
  • Chapter 5 The Happy Prince
  • Chapter 4 In the Kingdom of Fools
  • Email
  • Facebook
  • Pinterest
  • RSS
  • Twitter

Copyright © 2022 · About Us · Contact Us · Privacy Policy