Can the Uncertainty Principle Explain High Electron Energies in the Nucleus?

[Kazuya]

Hi, I'm a first year physics student at the University of Oregon and I was hoping that someone here might be able to help me with a problem I've been having.

A question on my assignment asks me to "Use the uncertainty principle to show that if an electron were present in the nucleus (r = approximately 10^-15 m), its kinetic energy (use relativity) would be hundreds of MeV. (Since such electron energies are not observed, we conclude that electrons are not present in the nucleus). [Hint: a particle can have energy as large as its uncertainty.]

Firstly, I decided to use the form of uncertainty principle (delta x)(delta p) is approximately larger than h-bar. I used (10^-15 m) as (delta x) and split (delta p) into m(delta v). I than used the rest mass of an electron (9.11x10^-31 kg) as m and attempted to solve for (delta v). [Using h-bar = 1.055x10^-34 J*s]. Unfortunately, this gave me an answer of (delta v) is approximately larger than (1.15x10^11 m/s)... which of course is much larger than c (3.00x10^8 m/s).

Already I knew there was a problem... when attempting to find K = (gamma - 1)mc^2, gamma becomes something like (382i)^-1/2. This is a definite problem because we haven't begun using non-real numbers (i) in this course yet.

Does anyone see an obvious problem with what I'm doing? Should I be using (delta E)(delta t) approximately larger than h-bar for the uncertainty principle? For my special relativity equation should I use E^2 =(p^2)(c^2) + (m^2)(c^4)?

I've spent several hours on this problem trying every imaginable route and have even come to the conclusion (incorrectly, of course) that [1+ (m^2)(c^2)] is approximately larger than 1/(2pi)^2! Any help is greatly appreciated.

 

[1100f]

Originally posted by Kazuya
Hi, I'm a first year physics student at the University of Oregon and I was hoping that someone here might be able to help me with a problem I've been having.

A question on my assignment asks me to "Use the uncertainty principle to show that if an electron were present in the nucleus (r = approximately 10^-15 m), its kinetic energy (use relativity) would be hundreds of MeV. (Since such electron energies are not observed, we conclude that electrons are not present in the nucleus). [Hint: a particle can have energy as large as its uncertainty.]

Firstly, I decided to use the form of uncertainty principle (delta x)(delta p) is approximately larger than h-bar. I used (10^-15 m) as (delta x) and split (delta p) into m(delta v). I than used the rest mass of an electron (9.11x10^-31 kg) as m and attempted to solve for (delta v). [Using h-bar = 1.055x10^-34 J*s]. Unfortunately, this gave me an answer of (delta v) is approximately larger than (1.15x10^11 m/s)... which of course is much larger than c (3.00x10^8 m/s).

Already I knew there was a problem... when attempting to find K = (gamma - 1)mc^2, gamma becomes something like (382i)^-1/2. This is a definite problem because we haven't begun using non-real numbers (i) in this course yet.

Does anyone see an obvious problem with what I'm doing? Should I be using (delta E)(delta t) approximately larger than h-bar for the uncertainty principle? For my special relativity equation should I use E^2 =(p^2)(c^2) + (m^2)(c^4)?

I've spent several hours on this problem trying every imaginable route and have even come to the conclusion (incorrectly, of course) that [1+ (m^2)(c^2)] is approximately larger than 1/(2pi)^2! Any help is greatly appreciated.

You began well by saying that p is of the order of hbar/Δx.
Instead of finding the velocity of the electron just compare pc with mc^2. Why that? Remember that E = γmc^2 and p = γmv. So that p/E = v/c^2 and if the velocity is of the order of c, you get E is approximately equal to pc. When you put this result in the equation E^2 =(p^2)(c^2) + (m^2)(c^4), you see thatthis means that you can neglect mc^2 wrt pc. In fact there is a terminology for this. A particle is said to be ultra-relativistic if its velocity is so closed to c that it can be treated as a zero mass particle where E=pc.

In the case of ultra-relativistic, almost all the energy will be kinetic energy.

If you look at your example you have p = h/r so that Ek = hbar*c/r.
remember that the compton wavelength of the electron is given by λ = hbar/mc, you will find that Ek = λ/r*mc^2.
For the electron, &lambda is about 10^-12 m, r is about 10^-15 and mc^2 is 0.5 MeV. So you get Ek is about 500MeV.

Hope I could help

 

[Kazuya]

Awesome =) Thanks for the help!