Heisenberg uncertainty principle and its application

Heisenberg uncertainty principle and its application


Heisenberg uncertainty principle and its application Is established by de-Broglie. L. De Broglie established the fact that all material particles in motion have wave like characteristics. This dual nature character gives rise to question . In this article we are going to discuss about Heisenberg uncertainty principle and its application .

Heisenberg uncertainty principle and its application

According to Heisenberg , It is impossible to determine simultaneously the position and momentum of electron with 100% accuracy.If, however , an experiment is designed for the exact determination of one of them the determination of other will become completely indeterminant or uncertain and vice versa.

Heisenberg uncertainty principle and its application

theory of Heisenberg uncertainty principle and its application

1) uncertainty in position and momentum-

Let Δx is uncertainty in position of a particle and ΔPx is uncertainty in momentum (in x-direction) then , the product of these can never be smaller then h/2 .

Mathematically, the uncertainty principle is written as

ΔxΔPx ≥  h/2

Where h is called Planck’s constant . Whose value is

h = h/2π = 6.62 Χ 10 ¯³⁴/ 2×3.14 = 1.054 × 10⁻³⁴

From above equation it is clear that if the momentum of particle is known accurately, the uncertainty in momentum is zero. I.e. ΔPx = 0 , then  Δx is infinite. Which means that According to uncertainty principle the particle is represented by a wave of constant amplitude extending over all space.In other words , the particle can be localised any where between −∞  to  +∞.  It is impossible to know both position and momentum of particle at the same time with accuracy.

In classical mechanics , h is zero i.e value of Planck’s constant is zero . However, in quantum mechanics Planck’s constant is not zero.

Heisenberg uncertainty principle and its application

2) uncertainty in energy and time-

According to Heisenberg , it is impossible that to determinen energy and time of a moving particle simultaneously,

∆E∆t ≥ h /2

3) uncertainty in Angel and momentum-

According to Heisenberg, it is impossible to determine angle and momentum of a moving particle simultaneously, with 100% accuracy.

∆θ∆J ≥ h /2

4) for microscopic particle uncertainty always →0

Experiment based on Heisenberg uncertainty principle

1) conceptual gamma ray microscope for the determination of position of an electron-

Heisenberg uncertainty principle and its application

When a photon of certain wavelength λ and initial momentum P (h/λ) after scattering enters into field of microscope. It may be anywhere within the angle 2θ . So that uncertainty in momentum is given by

∆P = [Psin θ ]₋θ+Θ

∆P = 2Psin θ

Also, ∆P =2h sinθ/λ

By resolving power of Microscope

∆x ≈ λ/2 sinθ

Now multiply above two equation

∆x∆P = 2h sinθ/λ . λ/2 sinθ

Hence,∆x∆P = h

∆x∆P ≈ h >h/2

If a particle is moving with velocity v and it is found in the distance ∆x then uncertainty in time ∆t,

∆t =∆x /v

Hence kinetic energy E = 1mv²/2

∆E = ∆(mv²/2)

Or, ∆E = mv∆v

∆E =v ∆(mv)

And,∆E =v ∆P

∆t∆E= (∆x /v ). (v ∆P)

∆t∆E= ∆x∆P> h/2

Diffraction of electron by a single slit –

Now as an another example let us consider the wave nature of a particle.

Heisenberg uncertainty principle and its application

A narrow beam of electrons of momentum p falls on a slit AB of the width ‘a’ . The de-Broglie wavelength assciatied with these moving electrons is equal to h/p.the electron passes through the slit,but the point at which it passes through the slit is not known.therefore the uncertainty in the position of the electron in the y-direction will be of the order of the width of the slit.

Condition of the diffraction minima –

First diffraction minima due to single slit

dsinθ =nλ


∆y = λ/sin θ

Uncertainty in momentum is

∆py= [psinθ]⁺θ₋θ


∆py= psinθ−(−psinθ)


By multiplying above two equations

∆py∆y=(  λ/sin θ).(2h sinθ/λ)

∆py∆y= 2h > h/4π

Application of Heisenberg uncertainty principle

1)Non existent of electron in nucleus :-

    According to Heisenberg principle,

∆x∆P ≥h/2 (1)

    Radius of the nucleus,
    ∆x ≈ 5× 10⁻¹⁵ m (2)
    Possibility of position of electron in nucleus
    Then from first equation ,

∆p =h/2 ∆x∆p =h/2(5×10⁻¹⁵m)

∆P= 1.054×10⁻³⁴/2×(5×10⁻¹⁵)

∆p≈ 1.054×10⁻²⁰kgm/sec

This is the minimum momentum of electron from mass energy relation-

E= √p²c²+m₀c⁴

Emin = √p²minc²+m₀c⁴

Rest,mass energy of electron is of order 0.511mev which is very less then p²minc²

p²minc²>> m₀²c⁴

    Emin = pmin c
    Emin= 1.054×10⁻²⁰×3 ×10⁸joule
    Emin= 20mev
    But β-decay experiment shows that emitted electron dosn’t have more then two theree mev energy that is electron cannot exist in nucleus.

2) ground state energy of harmonic oscillator-

Total energy of a particle which is oscillating SHM

    E=(mv²/2 )+ (kx²/2)
    E=(P²/2m) +(mω²x²)
     Minimum energy of oscillator,
    Emin= (∆P²/2m )+ (mω²(∆x²)/2)
    Put in equation
    Emin = (h/2∆x)²×1/2m + (mω²∆x²)/2

Emin=(h²/8∆x²m)+ mω²∆x²/2       (a)

    For minimum energy ,

dEmin /d ∆x =0 = d/d∆x (/8m∆x² + mω²(∆x²)/2)0=h/8m(-2/∆x³)+mω²(2∆x)/2

    In order to solve this equation we get
    ∆x²= h /2mω
    Put in equation(a)

Emin =  hω/2

3)natural width of spectrum lines-

    When an electron or atom transits from lower State to upper energy state, it remains 10⁻⁸Sec in the excited state and come back to the lower state after emitting E-M radiations. So that we have
    ,∆t = 10⁻⁸sec
    from Heisenberg uncertainty principle ∆E∆t ≈ h


    ∆E= 10⁻²⁶joule
    But we know


    ∆E = h∆v
    ∆v= ∆E/h
    ∆v = 10⁻²⁶/6.6×10⁻³⁴
    ∆v=1.5 ×10⁻⁷/sec
    That is frequency spreads and gives broad spectrum.

Heisenberg uncertainty principle and its application is highly successful in explaining various facts like dual nature of particle, non existance of electron in nucleus, state energy ground of harmonic oscillatgro ground state energy of hydrogen atom , natural width of spectrum lines etc.


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