Energy levels of partciles in a 3D box

[emilykay]

Energy levels of a particle in cubic box with sides length a are:

E= [(h^2 pi^2)/(2 m a^2)] * (nx^2 + ny^2 + nz^2)

where nx, ny, nz are integers 0.

If 10 electrons are placed in this box, what is the lowest possible total energy of all the electron?


.........

I am finding this question really intersting but I am stuck over one thing..
Why does it say "where nx, ny, nz are integers 0"

Surley this means that only 2 electrons can fit in the box, one spin up, one spin down?

I don't understand! please help...

 

[cristo]

Does is specifically say "where nx, ny, nz are integers 0" or is it some mathematical symbol that you have converted into words? (say, $n_x, n_y, n_z \in \mathbb{Z} - \{0\}$)

 

[emilykay]

It says "where nx, ny and nz are integers 0."

 

[Roman]

i think $$n_x=n_y=n_z=0$$ does mean, that all the electrons are in the ground state and as we know from Pauli's exclusion principle electrons (fermions) can only once occupy a quantum state which is specified by n and $$spin=\pm\frac{1}{2}$$. So as you noticed emilykay only 2 electrons can fit in the box, one spin up, one spin down

 

[emilykay]

Okay, thanks for your help!
It seems a bit strange though how this question is kindof a trick question seeing as its asking about 10 electrons. and its worth 6 marks so i don't know how it can be a trick...

Does anyone else know?

Emily.

 

[Roman]

Zero is not an integer.

you are right. the ground state energy is for n=1. maybe it should be $$n_x, n_y, n_z \not=0$$. so you should try all combinations and find the lowest possible energy

 

[emilykay]

It is...

 

[emilykay]

zero is an integer,

The lowest energy is (nx,ny,nz)=(0,0,0)

 

[Hootenanny]

Yes, you are right, sorry my bad.

 

[Roman]

The lowest energy is (nx,ny,nz)=(0,0,0)

then you have to start at 0

 

[emilykay]

thanks for your help. Definitly think its an error as the sentance doesn't really make sense.

 

[cristo]

I think it should read "where n_x, n_y, n_z are integers not equal to zero". Consider the one-dimensional case. If n=0 then $\psi$=0 everywhere, and so the particle is not in the box. This can then be generalised to the 3D case, and thus n_i must be non-zero.

 

[Jezuz]

I think the correct sentence is: "where n_x, n_y, n_z are integers not all equall to zero".

Usually you can have zero quanta in two directions for a particle. So what you need to do is to find the configuration of 10 particles that yields the lowest total energy. Remember that you must take into account the Fermi principle.

 

[pervect]

I would guess that there is a missing greater than sign in the problem, and that all of nx, ny, and nz have to be > 0.

Take a look at, for instance, http://en.wikipedia.org/wiki/Particle_in_a_box

While 0 is an integer, it's does not correspond to a valid solution. This can be deomnstrated by writing the appropriate expression for the wavefunction phi and imposing the boundary condition that phi=0 at the edge of the box.

 

[dextercioby]

The anwer to the problem is unique, if electron spin is taken into account. But there are several posibilities to take the triplet ($n_{1},n_{2},n_{3}$) to get the minimum energy.

Daniel.

 

[ccesare]

It should definitely read greater than 0.

The wavefunction for a three-dimensional particle in a box is just three one-dimensional wavefunctions multiplied together. Thus, a zero for any of the three quantum numbers reduces the total wavefunction to zero. Therefore, the lowest energy state is (1, 1, 1) and two electrons can fit here because of spin degeneracy. The next highest energy is (2, 1, 1), (1, 2, 1), and (1, 1, 2), with two electrons in each state. Continue in this way to place the last two electrons, but be careful to make sure that it is really the next level and not a higher one. E.g., which is lower, (1, 1, 3) or (2, 2, 1)?