Question

This is a strictly qualitative problem-no calculations allowed! Consider the "double square well" potential (Figure 2.21). Suppose the depth ${\mathit{V}}_{\mathbf{0}}$and the width a are fixed, and large enough so that several bound states occur.

(a) Sketch the ground state wave function ${\mathit{\Psi }}_{\mathbf{1}}$and the first excited state localid="1658211858701" ${\mathit{\Psi }}_{\mathbf{2}}$(i) for the case b = 0 (ii) for$\mathbf{b}\mathbf{\approx }\mathbf{a}$and (iii) for $\mathbf{b}\mathbf{\gg }\mathbf{a}$

(b) Qualitatively, how do the corresponding energies$\left(E_1\mathrm{and}{E}_2\right)$and vary, as b goes from 0 to ? Sketch ${\mathbf{E}}_{\mathbf{1}}\left(b\right)$and ${\mathbf{E}}_{\mathbf{2}}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$on the same graph.

(c) The double well is a very primitive one-dimensional model for the potential experienced by an electron in a diatomic molecule (the two wells represent the attractive force of the nuclei). If the nuclei are free to move, they will adopt the configuration of minimum energy. In view of your conclusions in (b), does the electron tend to draw the nuclei together, or push them apart? (Of course, there is also the internuclear repulsion to consider, but that's a separate problem.)

Short answer

(a)The sketch is

(i)

(ii)

(iii)

(b) The graph with and is,

(c)The electron forces the nuclei apart in the first excited state.

The ground state has the lowest energy in configuration I, and with b, the electron tends to attract the nuclei together, enhancing atom bonding.

Step-by-step solution

Define the energy

Power is generated by the use of physical or chemical resources, particularly to create light and heat or to operate machines.

Step 2: Draw the graph

(a)

Consider the potential graph shown below. If b=0 , the issue will be a standard finite square well with exponential decay outside and sinusoidal decay within. As a result of the potential's symmetry at the origin, the first solution is cosine, while the second is sine. There is only one node for the sine and none for the cosine.

1. For b=0

2. For $b\approx a$ Even ground condition. Outside, sinusoidal inside the wells, and hyperbolic cosine inside the barrier. The oddest initial excited state is a hyperbolic sine in the barrier. One without a node ${\psi }_1$ and two with a node ${\psi }_2$.

3. For b$\gg$ a Similar to (ii), the wave function in the barrier area is quite small. Essentially, there are two isolated finite square wells named and , which are even and odd linear combinations of the two distinct wells' ground states and are degenerate (in energy).

Step 3: Graph the values of E1(b) and E2(b)

(b)

When b is close to a in the second scenario, the exponential decay outside, sinusoidal inside the wells, and hyperbolic cosine inside the barrier all signify that the ground state is even. Odd, which in barrier terms denotes hyperbolic sine, describes the first excited state. There are no nodes for ${\psi }_1$and ${\psi }_2$, respectively.

Action of the electron

(c)

Here when b = 0 the energies above the bottom are shown:

$E_n+V_0\cong \frac{n^2{\pi }^2{\sout{h}}^2}{2m{\left(2a\right)}^2}$

so,

$E_1+V_0\approx \frac{{\pi }^2{\sout{h}}^2}{2m{\left(2a\right)}^2}{E}_2+V_0\approx \frac{4{\pi }^2{\sout{h}}^2}{2m{\left(2a\right)}^2}$

For $b\gg a$the width of each well is a,

$E_1+V_0\approx E_2+V_0\approx \frac{{\pi }^2{\sout{h}}^2}{2m{a}^2}$

Where:

$h=\frac{{\pi }^2{\sout{h}}^2}{2m{a}^2}$

The ground state has the lowest energy in configuration I, and with b, the electron tends to attract the nuclei together, enhancing atom bonding.

Therefore, the electron, on the other hand, forces the nuclei apart in the first excited state.