Electron Inside the Nucleus on Helium Atoms Problem

[Xeniebop]

Homework Statement

Quantum mechanics makes the electrons-in-the-nucleus theory untenable (for example on helium nucleus where before they know that a neutron exist, they thought that it was composed of four protons and 2 electrons). A confined electron is a standing wave, whose wavelength in the nucleus could be no longer than approx. 4R (R=radius of nucleus).

By analogy, the "fundamental" standing wave on a string has a wavelength satisfying L=1/2(lambda). In the case of the nucleus the "length of the string" is 2R.

Assuming a nuclear charge of 2e and a typical nuclear radius of R=5E-15. Show that the kinetic energy of an electron standing wave confined in the nucleus would be much greater then the magnitude of the attractive potential energy when the electron is at the surface of the nucleus?

And also I am confused about this, why at the surface of the nucleus? can the electron be found somewhere inside? and Is the electron moving fast or slow so I'm not sure if i have to consider relativistic effects.

Homework Equations

well here are the formulas I used (I don't know if they are correct)

For the kinetic energy of the electron
K=p^2/2m (p=momentum, m=mass)

to find the momentum I used

p=h/(lambda) (h=plank's, lamda=wavelength)

and to find lamda I used the given

1/2 lamda = 2 R thus, lamda=4R


To find the potential) i used (coulumb's law?)

U= 1/[4(pi)Eo] * (q1q2)/R

where q1=+2e (charge of the nucleus), q2=-e(charge of electron) and R=Radius of nucleus
(should i consider the contribution by the other electron, Helium has two electron right?)

The Attempt at a Solution

Well, I did the math and I indeed got K>U, kinetic energy is higher. But I'm not sure if my approach is correct ^^;

thanks ^^

 

[RafaMarce]

I think that the electron cannot be found in the nucleus, because the uncertainty principle will require a momentum uncertainty so high, so inside the nucleus it's worse. So i think you have to calculate the uncertainty in momentum of electron in nucleus surface, and use the relativistic energy E^2=(pc)^2+(mc^2)^2, to show aproximatelly that uncertainty in energy is so high compared with potencial energy.