Question

An electron is confined to a one-dimensional box. When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 1.2 eV. (a) What is the ground-state energy (in electronvolts) of the electron? (b) List all energies (in electronvolts) of photons that could be emitted when the electron starts in its second excited state and makes transitions downward to the ground state either directly or through intervening states. Show all these transitions on an energy level diagram. (c) What is the length of the box (in nanometers)?

Short answer

The ground-state energy of the electron is 0.4 eV. 2. What are the photon energies emitted when the electron transitions from the second excited state to the ground state? There are two possible transitions: a. Direct transition from n=3 to n=1 (E_3 -> E_1): The energy of the photon emitted in this transition will be [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 1^2) / (8 * m * L^2)]. b. Transition from n=3 to n=2 (E_3 -> E_2) and then from n=2 to n=1 (E_2 -> E_1): The energy of the photon emitted in this transition will be [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 2^2) / (8 * m * L^2)]. 3. How can the length of the box be calculated? The length of the box (L) can be calculated using the formula: L = sqrt((h^2) / (8 * m * E_1)) where h is Planck's constant, m is the mass of the electron, and E_1 is the ground-state energy.

Step-by-step solution

Calculate the energy difference between the first excited state and the ground state

We are given that the energy difference between the first excited state and the ground state (photon energy) is 1.2 eV. We can use the formula `E_n = (h^2 * n^2) / (8 * m * L^2)` to find this energy difference. For the first excited state (n = 2) and for the ground state (n = 1), the energy difference can be written as: `E_2 - E_1 = (h^2 * 2^2) / (8 * m * L^2) - (h^2 * 1^2) / (8 * m * L^2)` This energy difference is equal to 1.2 eV.

Calculate the ground-state energy

Now that we have an equation for the energy difference, we can solve it for the ground-state energy `E_1`. `E_1 = (h^2 * 1^2) / (8 * m * L^2) = (1/3) * (E_2 - E_1) = (1/3) * 1.2 eV` After the calculation, we get `E_1 = 0.4 eV`.

Calculate the photon energies emitted when the electron starts in its second excited state

We need to find the energies of the photons emitted when the electron transitions from the second excited state (n=3) to the ground state (n=1) either directly or through the first excited state (n=2). The possible transitions are: 1. Direct transition from n=3 to n=1 (E_3 -> E_1) 2. Transition from n=3 to n=2 (E_3 -> E_2) and then from n=2 to n=1 (E_2 -> E_1) We can use the energy formula to calculate the energies of these transitions: 1. `E_3 - E_1`: `E_3 = (h^2 * 3^2) / (8 * m * L^2)` The energy of the photon emitted in this transition will be: `E_3 - E_1 = [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 1^2) / (8 * m * L^2)]` 2. `E_3 - E_2`: `E_3 = (h^2 * 3^2) / (8 * m * L^2)` `E_2 = (h^2 * 2^2) / (8 * m * L^2)` The energy of the photon emitted in this transition will be: `E_3 - E_2 = [(h^2 * 3^2) / (8 * m * L^2)] - [(h^2 * 2^2) / (8 * m * L^2)]` The emissions for each transition can be represented on an energy level diagram.

Calculate the length of the box

We know the ground-state energy, E_1, and we have the expression for energy in terms of the length L: `E_1 = (h^2 * 1^2) / (8 * m * L^2)` We can solve for L: `L = sqrt((h^2) / (8 * m * E_1))` Given E_1 = 0.4 eV = 0.4 * 1.6 * 10^(-19) J and the mass of an electron, m = 9.11 * 10^(-31) kg, and Planck's constant, h = 6.63 * 10^(-34) J*s, plug in the values: `L = sqrt((6.63 * 10^(-34))^2 / (8 * 9.11 * 10^(-31) * 0.4 * 1.6 * 10^(-19)))` The length L will be in meters. To get the length in nanometers, multiply the result by 10^9. `L (nm) = L (m) * 10^9`