00:01
For this problem on the topic of quantum mechanics, we have a particle in three -dimensional box.
00:05
The y length of the box is twice the x -length, the z -length is a third of the y -length, and we want to know the energy difference between the first excited level and the ground level for the system.
00:14
We then want to know if the first excited level is degenerate, and in terms of the x -length, we want to know where the probability distribution will be the greatest in the lowest energy level.
00:24
Now, we want the energy difference between the first excited energy level and the ground state, and for this box, we have ly is 2lx, and we have lz equal to ly over 3, which we can write as 2lx over 3.
00:49
And so in terms of lx, we can write the energies as follows.
00:53
So, e, nx, n, n, and z is equal to nx, nx, and z, is equal to nx, nx, divided by lx squared plus n y squared over 4 l x squared plus 9 and z squared over 4 l x squared.
01:20
All of this multiplied by pi squared, h bar squared divided by 2m.
01:27
Now for the ground state, all of the quantum numbers equal to 1.
01:30
For the first exercise state, the smallest energy above the ground state is when nx is equal to n z is equal to 1 and n y is equal to 2...