Question

Here we consider adding two electrons to two "atoms," represented as finite wells. and investigate when the exclusion principle must be taken into account. In the accompanying figure, diagram (a) shows the four lowest-energy wave functions for a double finite well that represents atoms close together. To yield the lowest energy. the first electron added to this system must have wave function $\mathbf{A}$and is shared equally between the atoms. The second would al so have function $\mathbf{A}$and be equally shared. but it would have to be opposite spin. A third would have function B. Now consider atoms far a part diagram(b) shows, the bumps do not extend much beyond the atoms - they don't overlap-and functions $\mathbf{A}$and $\mathbf{B}$approach equal energy, as do functions $\mathbf{C}$and $\mathbf{D}$. Wave functions$\mathbf{A}$and$\mathbf{B}$in diagram (b) describe essentially identical shapes in the right well. while being opposite in the left well. Because they are of equal energy. sums or differences ofandare now a valid alternative. An electron in a sum or difference would have the same energy as in either alone, so it would be just as "happy" inrole="math" localid="1659956864834" $\mathbf{A}\mathbf{,}\mathbf{B}\mathbf{,}\mathbf{A}\mathbf{+}\mathbf{B}$, or$\mathbf{A}$- B. Argue that in this spread-out situation, electrons can be put in one atom without violating the exclusion principle. no matter what states electrons occupy in the other atom.

Short answer

${\mathrm{E}}_{1/2}=\frac{19{\mathrm{h}}^2}{8{\mathrm{mL}}^2},{\mathrm{E}}_1=\frac{5{\mathrm{h}}^2}{8{\mathrm{mL}}^2},{\mathrm{E}}_{3/2}=\frac{{\mathrm{h}}^2}{{\mathrm{mL}}^2}$

Step-by-step solution

Given information

$\begin{array}{cc}\text{State}& \text{Energy}\\ \mathrm{n}=1& {\mathrm{E}}_1=\frac{1^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\\ \mathrm{n}=2& {\mathrm{E}}_2=\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\\ \mathrm{n}=3& {\mathrm{E}}_3=\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\end{array}$

Concept of energy of particle in infinite square wall of length 

The energy of a particle in an infinite square well of length $\mathbf{L}$ is given by

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{n}}^{\mathbf{2}}{\mathbf{h}}^{\mathbf{2}}}{\mathbf{8}{\mathbf{mL}}^{\mathbf{2}}}$

Evaluate the energy of the particle in each state

Spin-1/2 Particles:

Since spin-1/2 particles are Fermions, the Pauli exclusion principle tells us that there can only be two particles in each state. So, in the minimum energy configuration, there will be two in the $n=1$state, two in the role="math" localid="1659957086918" $n=2$state, and one in the $\mathrm{n}=3$state.

The energy of a particle in an infinite square well of length$\mathrm{L}$is given by

${\mathrm{E}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

Thus, the energies of the particles in each state are

$\begin{array}{cc}\text{State}& \text{Energy}\\ \mathrm{n}=1& {\mathrm{E}}_1=\frac{1^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\\ \mathrm{n}=2& {\mathrm{E}}_2=\frac{2^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\\ \mathrm{n}=3& {\mathrm{E}}_3=\frac{3^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}\end{array}$

Therefore, the total energy for the system is

$\mathrm{E}=2{\mathrm{E}}_1+2{\mathrm{E}}_2+{\mathrm{E}}_3=\frac{19{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

Spin-1 Particles:

Since spin-1 particles are Bosons, they do not obey the Pauli exclusion principle. Thus, every one of the particles may be in the$n=1$ state. Therefore

$\mathrm{E}=5\cdot \frac{{\left(1\right)}^2{\mathrm{h}}^2}{8{\mathrm{mL}}^2}=\frac{5{\mathrm{h}}^2}{8{\mathrm{mL}}^2}$

Spin-3/2 Particles:

For spin-3/4 particles, there are four different possible value of ${\mathrm{m}}_{\mathrm{s}}:-\frac{3}{2},-\frac{1}{2},+\frac{1}{2},+\frac{3}{2}$.

Thus, without violation of the Pauli exclusion principle, four particles could have , with the fifth in the $\mathrm{n}=2$.

Therefore, total energy will be $\mathrm{E}=4{\mathrm{E}}_1+{\mathrm{E}}_2=8\cdot \frac{{\mathrm{h}}^2}{8{\mathrm{mL}}^2}=\frac{{\mathrm{h}}^2}{{\mathrm{mL}}^2}$