Applications of Heisenberg’s Uncertainty principle: Non-existence of electrons in the nucleus

Applications of Heisenberg Uncertainty principle

The Heisenberg uncertainty principle based on quantum physics explains a number of facts which could not be explained by classical physics. One of the applications is to prove that electron can not exist inside the nucleus. It is as follows:-

Non-existence of electrons in the nucleus

In this article, we will prove that electrons cannot exist inside the nucleus.

But to prove it, let us assume that electrons exist in the nucleus. As the radius of the nucleus in approximately 10-14 m. If electron is to exist inside the nucleus, then uncertainty in the position of the electron is given by

∆x= 10-14 m

According to uncertainty principle,

∆x∆px =h/2∏

Thus                            ∆px=h/2∏∆x

Or                               ∆px =6.62 x10-34/2 x 3.14 x 10-14

∆px=1.05 x 10-20 kg m/ sec

If this is p the uncertainty in the momentum of electron ,then the momentum of electron should be at least of this order, that is p=1.05*10-20 kg m/sec.

An electron having this much high momentum must have a velocity comparable to the velocity of light. Thus, its energy should be calculated by the following relativistic formula

E=  √ m20 c4 + p2c2

E =  √(9.1*10-31)2 (3*108)4 + (1.05*10-20)2(3*108)2

= √(6707.61*10-30) +(9.92*10-24)

=(0.006707*10-24) +(9.92*10-24)

= √9.9267*10-24

E= 3.15*10-12 J

Or                                  E=3.15*10-12/1.6*10-19 eV

E= 19.6* 106 eV

Or                                  E= 19.6 MeV

Therefore, if the electron exists in the nucleus, it should have an energy of the order of 19.6 MeV. However, it is observed that beta-particles (electrons) ejected from the nucleus during b –decay have energies of approximately 3 Me V, which is quite different from the calculated value of 19.6 MeV. Second reason that electron can not exist inside the nucleus is that experimental results show that no electron or particle in the atom possess energy greater than 4 MeV.

Therefore, it is confirmed that electrons do not exist inside the nucleus.

Heisenberg uncertainty principle

Statement: According to Heisenberg uncertainty principle, it is impossible to measure the exact position and momentum of a particle simultaneously within the wave packet.

We know, group velocity of the wave packet is given by

vg =∆ω/∆k

Where ω is the angular frequency and k is the propagation constant or wave number

But vg is equal to the particle velocity v

Thus vg = v =  ∆ω/∆k                                                   (1)

But                    ω=2пf

Where f is the frequency

Therefore   ∆ ω = 2п ∆ f                          (2)

Also                       k=2 п/λ

Since         de-Broglie wavelength   λ=h/p

By putting this value in equation of k, we get

k=2пp/ λ

Therefore                ∆k=2п∆p / λ                                 (3)

Put equations (2) and (3) in equation (1), we get

v= 2пh∆f/2п∆p =h∆f /            (4)

Let the particle covers distance ∆x in time ∆t, then particle velocity is given by

v  = ∆x/∆t                     (5)

Compare equations (4) and (5), we get

∆x/∆t=h∆f/∆p

Or                          ∆x.∆p=h∆f ∆t                                     (6)

The frequency ∆f is related to ∆t by relation

∆t≥ 1/∆f                                           (7)

Hence equations (6) becomes

∆x.∆p≥ h

A more sophisticated derivation of Heisenberg’s uncertainty principle  gives

∆x.∆p=h/2п                                          (8)

Which is the expression of the Heisenberg uncertainty principle.

As the particle is moving along x-axis. Therefore, the momentum in equation (8) of Heisenberg’s uncertainty principle should be the component of the momentum in the x-direction, thus equation Heisenberg’s uncertainty principle can be written as,

∆x.∆px=h/2п                             (9)

Note: There can not be any uncertainty if momentum is along y direction.

Q: Why there is uncertainty in position and momentum?

Answer: Because the particle is always in disturbed state during motion. It is not possible to calculate the position and momentum of particle simultaneously.