Quantum:- SHO potential and its energy

[indie452]

Homework Statement

the energy levels of a particle mass in a symmetric 3d SHO potential are:

E = (n_x + n_y + n_z + 3/2)*h-bar*[tex]\sqrt{\frac{C}{m}}[/tex]

C=constant
n=principal quantum number = n_x + n_y + n_z

A) If 10 electrons are in the potential what's the lowest possible value for the total energy of all the electrons?

B) If instead 10 pi^- mesons are placed in th same potential what is the lowest possible value of the total energy of the mesons?

The Attempt at a Solution

A)
for 10 electrons with + or - 1/2 spin the l quantum number is 2 as if the electrons were in an atom there would be s, p, d orbitals
This means that n is and integer greater than 2 [lowest n is 3]

so E = (3+3/2)*h-bar*[tex]\sqrt{\frac{C}{m}}[/tex])
me = mass of electron

= (9/2)*h-bar*[tex]\sqrt{\frac{C}{10me}}[/tex])

i did think that maybe the l quantum number should be 3 but it doesn't need to be in this example it just has 2 as the lowest possible. if all electrons had same spin direction then it would need l=3 due to PEP

B)
would the answer be the same as above? as l=2 but the mass would be 2730me as pi meson is approx 273 times mass of electron
= (9/2)*h-bar*[tex]\sqrt{\frac{C}{2730me}}[/tex])

 

[Matterwave]

Consider that the electron is a Fermion and the pi meson is a Boson. The energy levels they can occupy are different.

 

[vela]

The atomic quantum numbers [itex]nlm[/itex] don't apply at all here. That's for an electron in a Coulomb potential. In the 3D SHO, the states are labeled by [itex]n_x[/itex], [itex]n_y[/itex], and [itex]n_z[/itex]. The ground state would be [itex](n_x,n_y,n_z)=(0,0,0)[/itex]. At the next energy level, there are three degenerate states, [itex](1,0,0), (0,1,0), (0,0,1)[/itex], and so on.