Application of Heisenberg's Uncertainty Principle

[Aksaay]

]How can H.U.P be used to explain the following :
(i) The non-existence of the electron in the nucleus.
(ii) The existence of protons, neutrons and alpha particle.
(iii) The existence of finite zero point energy.
(iV) The binding energy of an electron in an Hydrogen atom is of the order of 15 eV.
(v) The radiation emitted from an excited atom does not appear with a precise frequency.

 

[Aksaay]

How can H.U.P be used to explain the non-existence of the electron in the nucleus.

 

[Drakkith]

Is there a reason you want the HUP to tell you this?

 

[Aksaay]

Well, i have a quantum assignment. I did lots of research and did not get a consistent answer.

 

[Drakkith]

I see. Well, if you have a question about homework, then I'd post in the homework section. That will get you the best chance of getting help on it.

 

[Aksaay]

I see. Well, if you have a question about homework, then I'd post in the homework section. That will get you the best chance of getting help on it.

Hehe ... i did so ... :P :P
no1 is interested in replyin' me there.. :P
so i posted it here...basically .. I didn't understand the concept that's y I posted it as a thread.

 

[Drakkith]

Ah I see. Honestly I don't know. I didn't think the HUP had anything to do with the electron not being in the nucleus, so if you have an assignment asking how it applies then I must be mistaken.

 

[martinbn]

Just a guess, the nucleus is very small and very heavy. So heavy that it can be considered (with a good approximation) as stationary. So if the electron were there, then you would have it in a small region and with a very small velocity.

 

[Aksaay]

hey buddy, i think i got an answer..But i don't know if its ok. Here it is :

The Heseinberg’s Uncertainty Principle states that you cannot know the position and momentum of a particle simultaneously. More rigorously stated, the product of the uncertainty of the position of a particle (Δx) and the uncertainty of its momentum (Δp) must be greater than a specified value:

∆x∆p ≥ (h/4π)

Now, as the electron approaches the nucleus, it's uncertainty in position decreases (if the electron is 10nm away from the nucleus, it could be anywhere within a spherical shell of radius 10nm, but if the electron is only 0.1nm away from the nucleus, that area is greatly reduced). According to the Heisenberg uncertainty principle, if you decrease the uncertainty of the electrons position, the uncertainty in its momentum must increase. This increased momentum uncertainty means that the electron will be moving away from the nucleus faster, on average.

Put another way, if we do know that at one instant, that the electron is right on top of the nucleus, we lose all information about where the electron will be at the next instant. It could stay at the nucleus, it could be slightly to the left or to the right, or it could very likely be very far away from the nucleus. Therefore, because of the uncertainty principle it is impossible for the electron to fall into the nucleus and stay in the nucleus.

In essence, the uncertainty principle causes a sort of quantum repulsion that keeps electrons from being too tightly localized near the nucleus.

 

[Aksaay]

Just a guess, the nucleus is very small and very heavy. So heavy that it can be considered (with a good approximation) as stationary. So if the electron were there, then you would have it in a small region and with a very small velocity.

Ok, if u say so.. But how can HUP account for the existence of protons, neutrons and alpha particle ?

 

[Drakkith]

Honestly, I don't see how the HUP applies to WHY the electron can't be in the nucleus. An electron can't occupy the nucleus because it simply cannot. Each electron is an a specific orbital with the appropriate propability map of it, which actually can have a small chance of an electron being INSIDE the nucleus. This doesn't mean it occupies the nucleus, like a proton would, but just that it has a small chance of being found there.

I don't see the HUP applying to this any more than I see it applying to a proton or neutron in the nucleus suddenly NOT being in the nucleus and occupying an orbital.

 

[ZapperZ]

Thread moved and merged.

Zz.

 

[Drakkith]

Would love for someone to explain how HUP explains all this, if it does.

 

[Dr_Zinj]

This question makes the assumption that the electron in question is bonded in an orbit about the nucleus, and not a free electron or beta particle.

An orbital electron shell is really a probability cloud layer. As an electron absorbs energy quanta, it gets boosted to higher energy shells. Geometrically, those shells should contain larger volumes (r1 and r2 from the center of the nucleus being greater than a lower shell, at least until you reach the heavier elements, and assuming the actually probability cloud remains the same depth at each layer.)

Now that's a thought; the fact that electrons in higher shells are more easily stripped away may indicate that the depth of the probability layer decreases. I wonder how I'd go about proving or disproving that by experimentation...?

 

[physicsamit]

The proof that electron can not exist inside the nucleus according to H.U.P. is that after taking assumption that if electron exist inside the nucleus and then doing calculation, we get that the energy of electrons come out very high. But that is experimentally not seen. The complete discussion and solution can be seen here:
http://www.winnerscience.com/quantum-physics/applications-of-heisenberg%E2%80%99s-uncertainty-principle-non-existence-of-electrons-in-the-nucleus/"