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

 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.

Once the beta decay occurs, the electron energy is determined by the difference of nuclear levels (conservation of energy). 

What you describe, is commonly calculated using the Fermi theory of beta decay, which starts by calculating the transition of the wave function of the nucleus into a system system where an electron becomes manifest within this nucleus.

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

 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.

The uncertainty principle states that ∆x∆px < h/2pi. So if we assume the nucleus is 1E-14 m, only electrons with energies above 19.6 MeV would become manifest in the nucleus. Electrons of lower energies can virtually exist within the nucleus without  contradicting the  uncertainty principle.  

 Thanks Erik and Saeed. 

The question is about whether the old uncertainty principle argument about non-existence of electrons inside the nucleus should be relied on or should it be discarded.

As Erik rightly points out, electrons with higher energies can be inside the nucleus, closer to the center of the nucleus, with an order of magnitude less position uncertainty, which will render the U.P. argument invalid. These electrons while being emitted from deep inside may upon doing work against the Coulomb attraction, finally appear to come out with lower energies as observed.

The uncertainty principle doesn't have anything to do with it. It's the result of Rutherford's experiment, as described in his paper, cf. http://www.chemteam.info/Chem-History/Rutherford-1911/Rutherford-1911.html

regarding the relative size of the distribution of positive charge vs. that of the distribution of negative charge, in the atom. 

The uncertainty principle relates the variances of conjugate variables; it doesn't imply anything about their average values, nor about the variances of the individual variables. So it's simply wrong to claim that the uncertainty principle can predict the energy of the electron emitted during beta decay. 

(incidentally, the uncertainty principle doesn't place an upper, but a lower, bound on the product of the variances of conjugate variables-assuming that the situation described is realized at all. So it can't be used to deduce the presence of electrons in the nucleus, since the presence is  assumed, in that case.)

Stam,

Do you mean to say all standard textbooks on modern physics (e.g. Beiser; Richtmeyer, Kennard, Cooper etc) got it all wrong when they introduced this example as an illustration/application/prediction/confirmation of the validity of the uncertainty principle?

Could you please elaborate more on your parenthetical remarks in the end?

I think, several things have been mixed up here.

The original argument (as I understand it) is that -because a nucleus is so small- a real electron in the nucleus would have a minimum of momentum which is higher than the momentum of beta-particles that can be observed. So, beta particles can not just be real electrons emitted from the nucleus that have been there before. So far, so good.

Secondly, the radial part of the electron density is not 0 at r=0 for all s electrons, so in fact there is a probability for s electrons to be found within the nucleus. Taking into account that the nucleus obtains a certain space, this is even true for p/d/f electrons (having electron density of 0 (only) at r=0 but having electron density>0 for any r>0, however small r may be).

Third, if energy can be gained from this process, a transition can occur in which an electron is generated within the nucleus and emitted because the possible electron wave functions in the nuclear (Coulomb) field are not zero.

So my personal opinion is that it is misleading to use the uncertainty principle to "demonstrate" that electrons can not exist in the nucleus. Vice versa: The uncertainty principle causes that electrons should not be thought as localised in the atom but rather have to be described in terms of electron densities / atomic orbitals. And these electron densities are not zero in the nucleus. 

I fully agree with you Eric. Thank you very much for your very lucid comments. Notes: 

>Secondly, the radial part of the electron density is not 0 at r=0 for all s electrons, so in fact there is a probability for s electrons to be found within the nucleus. Taking into account that the nucleus obtains a certain space, this is even true for p/d/f electrons (having electron density of 0 (only) at r=0 but having electron density>0 for any r>0, however small r may be).

> Here please note that the nucleus is treated as a point object in the solution of Schrodinger equation. So the radial wave function description of radial probabilities actually breaks down at the nuclear boundary. 

Best

@Rajat: Yes, in the solution of the Schrödinger equation, the nucleus is a point object. On the other hand, in Fermi theory the lepton wave functions (for l=0) are assumed to be constant  troughout the volume of the nucleus. So bringing this two point of views together, it is reasonable to assume some electron density in the nucleus if it is not assumed to be a point object.   

Very true Eric. This has been a very fruitful discussion. Thanks a lot.

 Thank you Richard for the comment. The question arose during discussions with undergraduate students in my QM classroom on the uncertainty principle argument for non-existence of electrons in the nucleus. The small energy difference really evokes suspicion about the acceptability of argument, unless one is somehow desperate to score a point in favor of UP, as was the case with Bohr and Heisenberg in the late 1920s. I think it is high time this particular example is taken off the textbooks and UG syllabi as a confirmation/application of UP.

Dear ALL,

Assuming that the UP argument is weak, can there be some other theoretical argument for the non-existence of the electron in the nucleus as its permanent constituent? 

In my opinion, why the electron cannot be found in the nucleus is because of the fact that it would immediately combine with the proton to become a neutron. There is a large cross-section  for this in the densely packed nucleus, if the electron somehow finds itself inside. In the days of Bohr-Heisenberg when the uncertainty principle was struggling to gain acceptance and when this possibility of e+p-->n+Neutrino was not known. (The neutron was yet to be discovered.). Thus the weak UP argument gained currency in the early part of the 1930's.

Learnt a lot from the discussion-especially helpful for Teachers.

Thank you sir for the kind remark.