The energy levels of an electron in a one-dimensional box are given by: E = (n^2 * h^2) / (8 * m * L — Physical Chemistry — Electrochemistry Chemistry Question
A frog in a well
The energy levels of an electron in a one-dimensional box are given by:
E = (n^2 * h^2) / (8 * m * L^2) n: 1, 2, 3…
in which h is the Planck’s constant, m is the mass of the electron, and L is the length of the box.
I. The π electrons in a linear conjugated neutral molecule are treated as individual particles in a one-dimensional box. Assume that the π electrons are delocalized in the molecular length with the total number of N π electrons and their arrangement is governed by the principles of quantum mechanics.
3.1 Derive the general expression for ∆ELUMO – HOMO when an electron is excited from the HOMO to the LUMO.
Model Answer
The general expression is given by:
∆ELUMO-HOMO = [(N/2 + 1)^2 * h^2 / (8mL^2)] - [(N/2)^2 * h^2 / (8mL^2)] = (N+1)h^2 / (8mL^2)
3.2 Determine the wavelength λ of the absorption from the HOMO to the LUMO.
Model Answer
From Planck’s quantum theory:
∆E = hc / λ
λ can be given by:
hc / λ = (N+1)h^2 / (8mL^2) → λ = (8mcL^2) / (h(N+1))
II. Apply the model of π electrons in a one-dimensional box for three dye molecules with the following structures (see the structural formula). Assume that the π electrons are delocalized in the space between the two phenyl groups with the length L is approximately equal to (2 k + 1)(0.140) nm, in which k is the number of the double bonds.
a) 1,4-diphenyl-1,3-butadiene (denoted as BD)
b) 1,6-diphenyl-1,3,5-hexatriene (denoted as HT)
c) 1,8-diphenyl-1,3,5,7-octatetraene (denoted as OT)
3.3 Calculate the box length L (Å) for each of the dyes.
Model Answer
For BD: L = (2×2 +1) 0.140 nm = 5×0.140 ⋅ 10^-9 m = 7 ⋅ 10^-10 m = 7.0 Å
For HT: L = (2×3 +1) 0.140 nm = 7×0.140 ⋅ 10^-9 m = 9.8 ⋅ 10^-10 m = 9.8 Å
For OT: L = (2×4 +1) 0.140 nm = 9×0.140 ⋅ 10^-9 m = 12.6 ⋅ 10^-10 m = 12.6 Å
3.4 Determine the wavelength λ (nm) of the absorption for the molecules of the investigated dyes.
Model Answer
From the general equation (3), the wavelength λ for each of the dyes are given:
BD: λ = 3.234 10^-7 m = 323.4 nm
HT: λ = 4.528 10^-7 m = 452.7 nm
OT: λ = 5.82 10^-7 m = 582.0 nm
III.
3.5 Recalculate the box length L (Å) for the three dye molecules, assuming that the π electrons are delocalized over the linear conjugated chain which is presented as a line plotted between the two phenyl groups (see the structural formula). The bond angle C–C–C is 120o and the average length of C–C bond is 0.140 nm.
Model Answer
The box length can be calculated based on the geometry of the C–C–C chain as follows:
The box length is a combination of a number of the length of 2d which is given by 2d = lC-C × sin 60° = (0.140 ⋅ 10^-9) × sin 60° = 1.21 ⋅ 10^-10 m.
Therefore, the box length for the three dye molecules can be calculated as follows:
For BD: L = 1.21 ⋅ 10^−10 m × 5 = 6.05 ⋅ 10^−10 m = 6.05 Å
For HT: L = 1.21 ⋅ 10^−10 m × 7 = 8.47 ⋅ 10^−10 m = 8.47 Å
For OT: L = 1.21 ⋅ 10^−10 m × 9 = 10.89 ⋅ 10^−10 m = 10.89 Å
IV. The following experimental data on the wavelength λ of absorption are given:
3.6 Determine the box length L (Å) of the linear conjugated chain for each of the three investigated dyes.
Model Answer
From equation (3): L = sqrt[(λ × h × (N+1)) / (8mc)]
For BD: L = 7.06 ⋅ 10^-10 m = 7.06 Å
For HT: L = 8.63 ⋅ 10^-10 m = 8.63 Å
For OT: L = 12.64 ⋅ 10^-10 m = 12.64 Å
3.7 Tabulate the values of the box length L for the dyes calculated above by the three different methods, denoted as I, II, and III.
Model Answer
The following table shows the values of the box length for the investigated dyes calculated with different methods:
3.8 Choose the method which is the most fit to the experimental data.
Model Answer
Method (3) is the best fit.