Question

An electron is confined to a one-dimensional box of length \(L\) (a) Sketch the wave function for the third excited state. (b) What is the energy of the third excited state? (c) The potential energy can't really be infinite outside of the box. Suppose that \(U(x)=+U_{0}\) outside the box, where \(U_{0}\) is large but finite. Sketch the wave function for the third excited state of the electron in the finite box. (d) Is the energy of the third excited state for the finite box less than, greater than, or equal to the value calculated in part (b)? Explain your reasoning. [Hint: Compare the wavelengths inside the box.] (e) Give a rough estimate of the number of bound states for the electron in the finite box in terms of \(L\) and \(U_{0}\).

Short answer

Answer: The potential energy outside the box in the finite case is finite, causing the wave function to decay exponentially rather than being zero outside the box. The energy of the third excited state for the finite box will be higher than that for an infinite box.

Step-by-step solution

Sketch the wave function for the third excited state

To sketch the wave function ψ(x) for the third excited state in a one-dimensional box, we must first find the wave function in the form ψ(x) = A sin(kx) + B cos(kx), where A and B are constants. Knowing that the wave function is zero at x = 0 and x = L, we find that ψ(x) = A sin(kx) because the cos(kx) term becomes zero at these boundary conditions. For the third excited state, we have n = 4 (since the ground state has n = 1). Using the formula k = n * pi / L, we find k = 4 * pi / L. Therefore, the wave function is ψ(x) = A sin(4 π x / L). To normalize the wave function, we integrate it from 0 to L, |ψ(x)|^2 dx = 1, and we find the value of A. So the wave function for the third excited state is ψ(x) = √(2/L) sin(4 π x / L).

Calculate the energy of the third excited state

To calculate the energy of the third excited state, we use the formula E = n²π²ħ² / (2mL²), with n = 4 for the third excited state, where ħ is the reduced Planck's constant, and m is the mass of an electron. Thus, the energy of the third excited state is E = 16π²ħ² / (2mL²).

Sketch the wave function for the finite box

For finite potential energy U₀ outside the box, the wave function ψ(x) will no longer be zero. It will decay exponentially outside the box, and the new wave function has the form of ψ(x) = C exp(-αx) + D exp(αx) for x < 0 and x > L. At x = 0 and x = L, the wave function must be continuous, so we have the same wave function as in Step 1 for the finite box: ψ(x) = √(2/L) sin(4 π x / L).

Comparing the third excited state energy between finite and infinite boxes

For a finite potential box, the potential energy outside the box is finite. Thus, the wave function cannot be zero outside the box, and it will decay exponentially. This behavior is observed in the wave function's sine and cosine terms, as these are the decaying exponential terms. As the potential energy U₀ outside the box becomes smaller, the wavelength inside also becomes smaller. This would lead to a higher energy E inside the box. Therefore, the energy of the third excited state for the finite box will be greater than that for an infinite box.

Estimate the number of bound states for the electron in the finite box

Let's estimate the number of bound states inside the finite box in terms of L and U₀. We know that the energy of the nth bound state is E = n²π²ħ² / (2mL²). The energy of the bound states will be lying between 0 eV and U₀. Thus, the maximum number of bound states n_max² * π²ħ² / (2mL²) = U₀. Solving for n_max, we obtain n_max = √(2mU₀L² / (π²ħ²)). Therefore, the approximate number of bound states in the finite box is given by n_max = √(2mU₀L² / (π²ħ²)). This expression gives us an idea of how many bound states we can expect inside the finite box.