Question

An electron in a one-dimensional box has ground-state energy 0.010 eV. (a) What is the length of the box? (b) Sketch the wave functions for the lowest three energy states of the electron. (c) What is the wavelength of the electron in its second excited state \((n=3) ?\) (d) The electron is in its ground state when it absorbs a photon of wavelength \(15.5 \mu \mathrm{m}\). Find the wavelengths of the photon(s) that could be emitted by the electron subsequently.

Short answer

The length of the box is approximately 378.5 nm. (b) How would you describe the wave functions for the three lowest energy states? 1. Ground state (n = 1): The wave function has only one half-wavelength fitting inside the box. 2. First excited state (n = 2): The wave function has a full wavelength fitting inside the box. 3. Second excited state (n = 3): The wave function has one and a half wavelengths fitting inside the box. (c) What is the wavelength of the electron in the second excited state? The wavelength of the electron in the second excited state is approximately 252.3 nm. (d) What is the wavelength of possible emitted photons after the electron absorbs a photon with a wavelength of 15.5 µm? The possible emitted photon would have a wavelength of approximately 2.766 µm.

Step-by-step solution

(a) Calculate the length of the box

Using the energy level formula for an electron in a one-dimensional box, we can rearrange terms to find the length L. The ground state energy level is \(E_1 = 0.010 \text{ eV}\). First, we need to convert 0.010 eV to Joules, using the conversion \({1 \text{ eV} = 1.602 \times 10^{-19} \text{J}}\): \(E_1 = 0.010 \text{ eV} \times \frac{1.602 \times 10^{-19} \text{J}}{1 \text{ eV}} = 1.602 \times 10^{-20} \text{J}\) Next, let's apply the energy level formula: \(E_1 = \frac{1^2 \pi^2 \hbar^2}{2m_edge L^2}\). Remember that \(\hbar = \frac{h}{2\pi}\), where \(h = 6.626 \times 10^{-34} \text{Js}\), and \(m_edge = 9.109 \times 10^{-31} \text{kg}\). Solving for the length L: \(L = \sqrt{\frac{1^2 \pi^2 \hbar^2}{2m_edge E_1}} = \sqrt{\frac{1^2 \pi^2 \frac{h^2}{(2\pi)^2}}{2m_edge E_1}} = \frac{h}{\sqrt{8m_edge E_1}} = \frac{6.626 \times 10^{-34}}{\sqrt{8 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-20}}}\) After calculating, we find \(L \approx 3.785 \times 10^{-10} \text{m}\) or \(378.5 \text{ nm}\).

(b) Sketch wave functions

The wave functions for the lowest three energy states are sine waves with the following formula for the n-th energy state: \(\psi_n(x) = \sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right)\) Remember that L is 378.5 nm. You can sketch the three states by considering the number of half-wavelengths contained in the box for each state: 1. Ground state (n = 1): The wave function has only one half-wavelength fitting inside the box. 2. First excited state (n = 2): The wave function has a full wavelength fitting inside the box. 3. Second excited state (n = 3): The wave function has one and a half wavelengths fitting inside the box.

(c) Wavelength in the second excited state

To find the wavelength of the electron in its second excited state, we can use the formula \(\lambda = 2L/n\). Replacing the values for L and n, we get: \(\lambda = \frac{2 \times 378.5}{3} \approx 252.3 \text{ nm}\)

(d) Wavelengths of emitted photons

The electron absorbs a photon of wavelength \(15.5 \mu \mathrm{m}\) in its ground state. Let's find the absorbed energy, using the formula \(E_\text{photon} = \frac{hc}{\lambda_\text{photon}}\). Absorbed energy \(= E_\text{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{15.5 \times 10^{-6}} \approx 1.279 \times 10^{-19} \text{J}\) Now add first energy level: \(E_\text{total} = E_1 + E_\text{photon} = 1.602 \times 10^{-20} + 1.279 \times 10^{-19} = 1.439 \times 10^{-19} \text{J}\) Calculate the energy difference between energy levels (\(\Delta E =E_\text{total} - E_n\)) and divide by Planck's constant (\(\frac{\Delta E}{h} = \frac{c}{\lambda_\text{emitted}}\)). Energy difference: \(\Delta E_{31} = 1.439 \times 10^{-19} - 9 \times 1.602 \times 10^{-20} = 7.18 \times 10^{-20} \text{J}\) Next, we calculate the wavelength of the emitted photon: \(\lambda_\text{emitted} = \frac{hc}{\Delta E_{31}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{7.18 \times 10^{-20}} \approx 2.766 \times 10^{-6} \text{m} = 2.766 \mu \mathrm{m}\) So, the photon(s) that could be emitted by the electron after absorbing the energy would have a wavelength of 2.766 \(\mu\)m.