The electronic structure of condensed matter is usually different from that of an isolated atom. For — Physical Chemistry — Kinetics Chemistry Question
Electronic structure of condensed matter
The electronic structure of condensed matter is usually different from that of an isolated atom. For example, the energy levels of a one-dimensional (1D) chain of Na atoms are illustrated in Figure 1. Here, the energy-level changes of the 3s-derived states of Na are shown. The energy-level spacing decreases as the number of Na atoms (N) increases. At an extremely large N, the energy-level spacing becomes negligibly smaller than the thermal energy, and the set of 3s-derived levels can be considered as a “band” of energy levels (last image in Figure 1). Na 3s electrons occupying the band of energy levels delocalize over the chain leading to a metallic character. Therefore, the 3s electrons can be assumed to be free particles confined in a 1D box.
The eigenenergy of the free particles confined in a 1D box is described as
E = n^2 h^2 / 8 m L^2 (n = 1, 2, 3, ···)
where n is the quantum number, h is the Planck constant, m is the weight of the electron, and L is the length of the 1D Na chain. Assuming that the chain length L = a0(N – 1), where N is the number of Na atoms and a0 is the nearest-neighbor interatomic distance, calculate the energy of the highest occupied level.
Model Answer
Since one eigenstate is occupied by two electrons with opposite spin directions (up-spin and down-spin), the quantum number n of the highest occupied level is N/2 for even N and (N+1)/2 for odd N. The length of the chain L is expressed as a0(N–1). On the basis of the given eigenenergy, the energy of the highest occupied level is written as
E = N^2 h^2 / 32 m a0^2 (N–1)^2 for even N and
E = (N+1)^2 h^2 / 32 m a0^2 (N–1)^2 for odd N.
We assume that 1.00 mg of Na forms a 1D chain with a0 = 0.360 nm. Calculate the energy width from the lowest occupied level to the highest occupied level.
Model Answer
The number of Na atoms present in 1.00 mg of Na is
N = (1.00×10^-3 / 23.0) × 6.02×10^23 = 2.617×10^19.
The energy width is expressed as
E(N/2) - E(1) = h^2 / 32 m a0^2 { (N^2 - 4) / (N - 1)^2 } for even N
and E((N+1)/2) - E(1) = h^2 / 32 m a0^2 { ((N+1)^2 - 4) / (N - 1)^2 } for odd N.
Since N is extremely large, the energy width is calculated as
h^2 / 32 m a0^2 = 1.16×10^-19 J for the both cases.
If the thermal energy at room temperature is assumed to be 25 meV, how many Na atoms are required when the energy gap between the highest occupied level and the lowest unoccupied level is smaller than the thermal energy (25 meV)? Calculate the least number of Na atoms required assuming that the number is even.
Model Answer
The energy gap for even N is ∆E = E(N/2 + 1) - E(N/2); using the equation obtained in (a), the equation for energy gap is rewritten as
∆E = h^2 / 32 m a0^2 { ((N+2)^2 - N^2) / (N-1)^2 } = h^2 / 8 m a0^2 { (N+1) / (N-1)^2 }.
We solve the equation h^2 / 8 m a0^2 { (N+1) / (N-1)^2 } = E_Thermal (25 meV).
This equation is rewritten as (N-1)^2 / (N+1) = h^2 / 8 m a0^2 E_Thermal = 116.2.
Thus, we obtain the quadratic equation N^2 – 118.2 N – 115.2 = 0, and solve the equation to obtain N = 119.2.
Therefore, at least 120 Na atoms are required when the energy gap is smaller than the thermal energy 25 meV.