Question

Heisenberg uncertainty principle existence of protons in nucleus mathematical proof

Answer 1

According to uncertainty principle,

∆x∆p =h/2*pi

Thus ∆p=h/2*pi*∆x

Or ∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=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 which is 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.

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Answer 2

The uncertainty principle is as follows:

∆x∆p =h/2*pi

∆p=h/2*pi*∆x

∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=1.05 x 10-20 kg m/ sec

Every atom is neutral as it has an equal number of protons and an equal number of electrons.

In the event an electron does not collide with the nucleus despite the presence of an electrical attraction can be said to occur due to the Heisenberg uncertainty principle, confining the electron’s position with respect to the nucleus causes it to have a large momentum uncertainty which as a result it will be said to ‘fly away. Protons like all others, are also under the effect of the same Heisenberg uncertainty principle but seem to stay in the small nucleus space, unlike an electron.

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