# Uncertainty Principle and the size of an atom

1. Jul 11, 2009

### nobahar

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 travelling 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.

2. Jul 11, 2009

### Naty1

m (delta(v))(delta(x)) greater than h, says it all...

3. Jul 11, 2009

### gabbagabbahey

First, it is incorrect to say that the electron will be travelling 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: Jul 11, 2009
4. Jul 11, 2009

### dave_baksh

gabba> you mean v subscript p in your last line of working

5. Jul 11, 2009

### gabbagabbahey

Yes, thank you. I've edited my post.

6. Jul 13, 2009

### nobahar

Thanks Gabba,
That's extremely clear and awesome.
Thanks!