Question

A cubical box of widths L_x = L_y = L_z = L = 6.0 nm contains three electrons. What is the energy of the ground state of this system? Assume that the electrons do not interact with one another, and do not neglect spin.

Answer 1

The energy of the single particle states in the cube I of side L is
  
So, the energies of the first few single particle energy states are in the unit of
   , i.e.,

   

Serial no.					Energy states
1	1	1	1	3	Ground state
2	1	1	2	6	1st excited state
3	1	2	1	6
4	2	1	2	6

And so, if we include the spin of the electron, then, there are two possible spins. So, the ground state energy of three electron system is given by
  

Electron			Spin of the electron	Total energy
1st	1, 1, 1	3	Spin up	3 + 3 + 6 = 12
2nd	1, 1, 1	3	Spin down
3rd	1, 1, 2	6	Spin up

So, the total energy of the ground state of the three electron system is given by
  
And for electron, . And for given , we have