The π-electrons of the iron–heme of a hemoglobin molecule can be visualized as a system of free elec — Physical Chemistry — Electrochemistry Chemistry Question
A Particle in a 2–D Box Quantum Mechanics
The π-electrons of the iron–heme of a hemoglobin molecule can be visualized as a system of free electrons moving in a two–dimensional box. According to this model, the energy of the electron is limited to the values:
E_{n_x, n_y} = (h^2 / 8 m_e L^2) * (n_x^2 + n_y^2) (n_x, n_y = 1, 2, 3, . . . .)
where h = 6.63×10^–34 J s is the Planck constant; n_x and n_y are the principal quantum numbers; m_e = 9.11×10^–31 kg is the electron mass; L is the length of the box.
Construct an energy level diagram showing the relative ordering of the lowest 17 orbitals.
Model Answer
E1,1 = 2 E0
E1,2 = E2,1 = 5 E0
E2,2 = 8 E0
E1,3 = E3,1 = 10 E0
E2,3 = E3,2 = 13 E0
E1,4 = E4,1 = 17 E0
E3,3 = 18 E0
E2,4 = E4,2 = 20 E0
E3,4 = E4,3 = 25 E0
E1,5 = E5,1 = 26 E0
where E0 = h^2 / 8 me L^2
Given the molecule contains 26 electrons, determine the electron population of the highest occupied orbitals in the ground state.
Model Answer
The total number of electrons in the highest occupied energy level is 4.
Assuming Hund's rule can be applied to this system, predict whether or not this system is paramagnetic.
Model Answer
Ground state is diamagnetic.
Light is absorbed only when the condition hν = ΔE is satisfied. If the length L for this 2D box is 1 nm, what is the longest wavelength of light that can lead to excitation? Express your result in nm. [The speed of light, c = 3.00×10^8 m s^–1.]
Model Answer
The longest–wavelength excitation energy is ΔE = (25 – 20) E0, where
E0 = (6.63×10^–34 J s)^2 / [8 × 9.11×10^–31 kg × (1×10^–9 m)^2] = 6.02×10^–20 J
ΔE = (25 – 20) E0 = 3.01×10^–19 J
The wavelength is
λ = h c / ΔE = [6.63×10^–34 J s × 3×10^8] / 3.01×10^–19 = 660 nm