Uncertainty principle for electron inside atom

[leehufford]

Homework Statement

Hopefully it is not a faux pas to post two questions relatively close together in time. This is more of a conceptual question than a calculation based question.

An electron is confined to a region of space of the size of an atom (0.1 nm). a) What is the uncertainty in the momentum of the electron? b) What is the kinetic energy of an electron with momentum equal to delta p? c) Does this give a reasonable value for the kinetic energy of an electron inside an atom?

Homework Equations

(delta x)(delta p) ~ (h bar)

The Attempt at a Solution

a) (h bar)/(delta x) ~ delta p = (1.05e-34 Js)/(0.1 nm) = 1.05e-24 kgm/s.
b) delta p = sqrt(2mK). K = 6.05e-19 J.
c) I don't know.

I feel like the uncertainty in momentum is very small, which is odd, because I thought the uncertainty in x was very small...maybe I don't have the proper perspective here. When either delta x or delta p are small, the other should be large. Also, I have no basis for whether this 6.05e-19 J, which should be 3.78 eV, is reasonable or not. Is it "supposed" to be reasonable, or is the point of this question to discover that it is not reasonable? Thanks in advance,

Lee

 

[mm8070]

The uncertainty is delta x delta p >= (hbar)/2 so you want (h bar)/(2*delta x) ~ delta p = (1.05e-34 Js)/(2*(0.1 nm)) = 5.5e-25. Second, consider the possibilities of momentum in that small amount of space. The wavelength of the electron would have to be less than 0.1nm so the smallest momentum would be p=h/0.1nm=6.626e-24. Or p = (6.626 + or - .55) e-24. This seems like a decent uncertainty in momentum to me because the uncertainty will be at most 8.3% of the momentum.

As for a reasonable KE, we already showed that the smallest the momentum can be within the 0.1nm atom is 6.626e-24. A momentum that is more than 10 times (i.e. 5.5e-25) less implies that the KE will be 100 times less than what it should be within that space of the atom.

Let me know if you see any flaws in my calculations or logic. It's been a couple of years since i was in quantum.

 

[leehufford]

Thank you for the response. In our book, the reasoning goes that since the uncertainty is >= h bar/2, we will almost certainly do worse than h bar/2, so our book says that delta p delta x ~ h bar is a reasonable estimate, and at the end of the chapter it lists this formula, with the (1/2) dropped from the equation. Do the results imply that a 0.1 nm uncertainty in position is a large uncertainty?

As for your logic, I followed the first paragraph but sort of got lost in the second one. Are you saying the KE is reasonable, or not so much? Thanks again,

Lee

 

[mm8070]

Considering the size of an electron compared to an atom, yes, an uncertainty of 0.1nm (about ten times the diameter of one hydrogen atom) is very large.

As for KE: I believe I'm correct in calculating the smallest possible momentum. If you compare the momentum uncertainty to the smallest momentum, you'll see that the uncertainty is more than 10 times less. If you were to use the momentum uncertainty to calculate KE_u and compare it to the smallest momentum to calculate KE_s, then you'll see that the KE_u is more than 100 times less than KE_s. Meaning the smallest KE possible is still 100 times more than KE_u. Meaning, with the given uncertainty, KE_u is physically impossible.

 

[leehufford]

0.1 nm is one angstrom. I think you have your units backwards there.

 

[mm8070]

Ah, you're right. The diameter of hydrogen is about 0.1nm. Even so, an electron is very small compared to the size of one hydrogen atom. I would still assert that 0.1nm is a large uncertainty for x.

 

[leehufford]

Thanks again for the reply. I agree than 0.1 nm is large compared to a basically point object (electron). I'm not sure where you are getting terms like "smallest kinetic energy" from. If the KE is unreasonable, would you happen to know what a common electron energy inside an atom is? I know its different for different energy levels and different atoms, but if my KE is unreasonable it should be very different from the average or sample values.

Lee

 

[unscientific]

I think to see whether it's reasonable we need to compare it to the energy levels of the hydrogen atom - it's 13.6eV for ground state. How does your answer compare to that?

 

[Orodruin]

What is a "large" uncertainty in position is quite arbitrary. The main thing to notice is that $\hbar$ is a very small number when expressed in Js so if you are using SI units you will get small numbers no matter what.

Based on the virial theorem, the kinetic energy in a bound state should equal the absolute value of the energy of the state (note that the bound state energy is typically taken as being negative with unbound states having positive energy).