Question

Consider an electron that is confined to a one-dimensional infinite potential well of width \(a=0.10 \mathrm{nm},\) and another electron that is confined to a three-dimensional (cubic) infinite potential well with sides of length \(a=0.10 \mathrm{nm}\). Let the electron confined to the cube be in its ground state. Determine the difference in energy ground state of the two electrons and the excited state of the one-dimensional electron that minimizes the difference in energy with the three-dimensional electron.

Short answer

Answer: The excited state for the 1D electron that minimizes the difference in energy with the 3D electron has a quantum number of 2 (n'=2).

Step-by-step solution

Finding the energy ground state for 1D electron

We have a one-dimensional infinite potential well of width \(a = 0.10\) nm, and for the lowest energy ground state, we set the energy quantum number \(n = 1\). To find the energy for this electron, we plug these values into the 1D energy formula: \(E_{1D} = \frac{(1)^2 \pi^2 \hbar^2}{2m(0.10 \times 10^{-9})^2}\) Calculating the energy, we get: \(E_{1D} = 6.024\times10^{-20} J\)

Finding the energy ground state for 3D electron

The electron is confined to a three-dimensional cubical well with sides of length \(a = 0.10\) nm. For the lowest energy ground state, we set the energy quantum numbers \(n_x = n_y = n_z = 1\). Plugging these values into the 3D energy formula: \(E_{3D} = \frac{(1^2+1^2+1^2)\pi^2\hbar^2}{2m(0.10\times10^{-9})^2}\) Calculating the energy, we get: \(E_{3D} = 1.807\times10^{-19} J\)

Determine the difference in energy ground states

Now, we will find the difference in energy ground states between the 1D and 3D electrons: \(\Delta E = E_{3D} - E_{1D} = 1.807\times10^{-19} J - 6.024\times10^{-20} J\) \(\Delta E = 1.205\times10^{-19} J\)

Finding the excited state for 1D electron

We need to find the excited state (\(n'\)) of the 1D electron which minimizes the difference in energy with the ground state energy of the 3D electron. Using the energy formula for the 1D electron: \(E'_{1D} = \frac{{n'}^2\pi^2\hbar^2}{2ma^2}\) We need to minimize the difference: \(\Delta E = |E_{3D} - E'_{1D}| = |1.807\times10^{-19} J - \frac{{n'}^2\pi^2\hbar^2}{2m(0.10\times10^{-9})^2}|\) To minimize the difference, we need a value of \(n'\) such that: \(n'^2 = \frac{2m(0.10\times10^{-9})^2}{\pi^2\hbar^2}(E_{3D}-E_{1D})\) Plugging the values of energy from previous steps, and calculating \(n'\): \(n' = \sqrt{\frac{2(9.109\times10^{-31}\,\text{kg})(0.10\times10^{-9}\,\text{m})^2}{\pi^2(1.055\times10^{-34}\,\text{J s})^2}(1.205\times10^{-19}\,\text{J})} \approx 1.87\) As we can't have fractional quantum numbers, we round up \(n'\) to the nearest integer. So, the closest excited state for the 1D electron is: \(n' = 2\) The strategy to find the excited state involves finding the \(n'\) value that would make the energy difference between the 1D and the 3D electron as close as possible. This allowed us to determine that an excited state with a quantum number of 2 would minimize this difference.