Why electrons can't remain in the nucleus

In summary, the conversation discusses the momentum and speed of an electron confined to an atomic nucleus, taking into account the uncertainty principle. The formula for the speed of the electron is derived and compared to the escape velocity, leading to the conclusion that electrons cannot remain in the nucleus due to their high speeds. The conversation also touches on the topic of special relativity and the importance of considering the changing mass of the electron at high speeds.
  • #1
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Homework Statement


Estimate the momentum of an electron confined to an atomic nucleus of radius a, and deduce a formula for its speed. By comparing the speed to the escape velocity explain why electrons cannot remain in the nucleus.


Homework Equations


Uncertainty principle


The Attempt at a Solution


Uncertainty Principle: ∆x∆p≈h (1)

If the electron were confined to the nucleus, the uncertainty in its position ∆x would be the radius of the nucleus. This means that the uncertainty in its momentum is given by rearranging (1) to:

∆p= ∆(mv)≈h/∆x => ∆v≈h/(m ∆x) (2)

The uncertainty in the velocity ∆v is all we know about the velocity so this is the velocity that the electron may have. (Is this explained right? How may I word this better – I get confused between uncertainty and actual velocity.)

∆v≈(6.626×10^(-34))/(9.11×10^(-31)×10^(-15) )=(6.626×10^12)/9.11

= 7.3×10^11 (this is faster than speed of light so must be wrong?)

Potential Energy of an electron in the nucleus of a Bohr atom is PE= e^2/(4πε_0 a) where the radius of the nucleus is a~10^(-15) metres.

Putting in values => PE=2.3×10^(-13) Joules

Escape velocity can be calculated from KE = PE.

KE=(m_e v^2)/2=PE
=> v = √(2PE/m_e ) =√((2 ×2.3×10^(-13))/(9.11×10^(-31) ))=707 106 781
=> v ≈ 7×10^8 ms^(-1)

So Escape Velocity ≈7×10^8 ms^(-1)

This is faster than the speed of light so I must have done something wrong as this would imply that the electron cannot escape unless its velocity is greater than the speed of light, therefore the electron would be bound?
 
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  • #2
Hi there,

From a first glance at your calculus, you do not take special relativity into it. Therefore, into classical physics, many problems can lead to solution where the speed of the object is greater than the speed of light.

Cheers
 
  • #3
Hi, many thanks for your comment, however I don't think the question intends special relativity to be used - it's in part of a course on QM. Not sure if that's the right attitude to take?! However, can you see any QM related things I have missed or done wrong?
 
  • #4
Hi there,

Not being an expert in QM, and having very bad memories from my classes in University, I don't see anything wrong with your equation, from QM point of view.

Eventhough, since you are talking about speed that are fractions of the speed of light, you anyways need to uses special relativity.

The problem that I see with you equation is that you suppose the mass of the electron constant, which is not true for speed that great.
∆p= ∆(mv)≈h/∆x => ∆v≈h/(m ∆x) (2)
is not right from that point of view. It should be something like:
[tex]\Delta p = \Delta (mv) \approx \frac{h}{\Delta x}[/tex]
Since the mass cannot be considered constant, this becomes:
[tex]\Delta v \approx \frac{h}{\Delta m \cdot \Delta x} [/tex]

Cheers
 
  • #5
Many thanks for your suggestions - I am not sure how your suggestion would lead to a different answer - in other words I'm still confused!
 
  • #6
Hi there,

I am just pointing out that electrons relevant speed in atoms must be viewed as relativistic particles. Therefore, the electron's mass is not it's rest mass anymore.

Cheers
 
  • #7
In relativity, momentum is defined as
[tex]p = \frac{mv}{\sqrt{1 - v^2/c^2}}[/tex]
and kinetic energy is defined as
[tex]K = \left(\frac{1}{\sqrt{1 - v^2/c^2}} - 1\right)mc^2[/tex]
where in each case m is what used to be called the "rest mass" (which is constant) not the "relativistic mass" (which changes with velocity).

As for explaining how to get the velocity: you could say that an electron stuck in the nucleus, over time, has an average velocity of 0 (otherwise it wouldn't be stuck). But if the uncertainty in that velocity is ∆v, it means the electron would actually have a velocity in some range whose characteristic width is ∆v - in some sense, the "average" magnitude of velocity is ∆v (for some definition of average).
 

1. Why can't electrons remain in the nucleus?

Electrons cannot remain in the nucleus because they have a negative charge and are repelled by the positively charged protons in the nucleus. This repulsion causes the electrons to orbit around the nucleus at specific energy levels.

2. What is the role of the strong nuclear force in preventing electrons from remaining in the nucleus?

The strong nuclear force is responsible for holding the positively charged protons together in the nucleus. However, it is a short-range force and is not strong enough to hold the negatively charged electrons within the nucleus.

3. How do electrons maintain their stability while orbiting the nucleus?

Electrons maintain their stability while orbiting the nucleus due to the balance between their kinetic energy and the attractive force of the positively charged nucleus. This balance keeps the electrons in a stable orbit without falling into the nucleus.

4. Can electrons ever enter the nucleus?

No, electrons cannot enter the nucleus as it goes against the laws of physics. The energy levels of the electrons are quantized, meaning they can only exist at specific distances from the nucleus. And since the nucleus is much smaller than the orbit of electrons, it is impossible for them to enter it.

5. What happens if an electron is forced into the nucleus?

If an electron is forced into the nucleus, it would result in a highly unstable and energetic state. This would cause the atom to undergo a nuclear reaction, resulting in the release of large amounts of energy. This process is known as electron capture and is a rare occurrence in nature.

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