Uncertainty Principle and the size of an atom

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The discussion centers on the Heisenberg uncertainty principle and its implications for the speeds of protons and electrons when confined to the same volume. It clarifies that the electron's uncertainty in speed is approximately 2000 times greater than that of the proton, not that it travels 2000 times faster. By confining both particles to the same space, their position uncertainties are equal, leading to a relationship between their momentum uncertainties. The calculations show that the electron's change in speed is proportional to the mass ratio of the proton to the electron. This explanation effectively illustrates the consequences of the uncertainty principle in particle physics.
nobahar
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Hey,
Sorry, but I have a qustion on the uncertainty principle to join the many others.
Just reading a book on physics, and its says that, as a result of the Heisenberg uncertainty principle, if the proton and electron were confined to the same volume of space, the electron would be traveling about 2,000 times faster, as the proton is 2,000 times bigger. How is this a consequence of the uncertainty principle? It must entail the momentum and the position, but I don't see how.
Thanks in advance for your responses.
 
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if the proton and electron were confined to the same volume of space, the electron would be traveling about 2,000 times faster, as the proton is 2,000 times bigger.

m (delta(v))(delta(x)) greater than h, says it all...
 
First, it is incorrect to say that the electron will be traveling 2000 times faster; the correct statement would be that the uncertainty in the electron's speed is about 2000 times greater than the the uncertainty in the proton's speed.

By confining the proton and the electron to the same volume, you are essentially saying that the uncertainty in position is the same for both particles (i.e. \Delta x_{\text{electron}}=\Delta x_{\text{proton}} )

So, if you assume that \Delta x_e\Delta p_e=\frac{\hbar}{2} and \Delta x_p\Delta p_p=\frac{\hbar}{2} then you have:

\Delta x_e\Delta p_e=\Delta x_p\Delta p_p

\implies \Delta p_{\text{electron}}=\Delta p_{\text{proton}}

Then you simply use the fact that \Delta p_e=m_e\Delta v_e and that \Delta p_p=m_p\Delta v_p (since you presumably know the masses of the electron and proton exactly, the uncertainty in their momenta is due entirely to the uncertainty in their speeds) and you get

\Delta v_e =\frac{m_p}{m_e} \Delta v_p \approx 2000\Delta v_p
 
Last edited:
gabba> you mean v subscript p in your last line of working
 
dave_baksh said:
gabba> you mean v subscript p in your last line of working

Yes, thank you. I've edited my post.
 
Thanks Gabba,
That's extremely clear and awesome.
Thanks!
 
I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question. Ex. 1 - A moving object impacts a static object at a constant velocity. Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact. Assuming the mass of the objects is the same and the velocity at the moment of impact...

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