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Chapter 2: Problem 17

A very simple model of a polyene is the free electron molecular orbital (FEMO) model. Regard a chain of \(N\) conjugated carbon atoms, bond length \(R_{\mathrm{CC}},\) as forming a box of length \(L=(N-1) R_{\mathrm{CC}} .\) Find an expression for the allowed energies. Suppose that the electrons enter the states in pairs so that the lowest \(\frac{1}{2} N\) states are occupied. Estimate the wavelength of the lowest energy transition, taking \(R_{\mathrm{CC}}=140 \mathrm{pm}\) and \(N=22 .\) Repeat the calculation of the wavelength if the length of the chain is taken to be \((N+1) R_{\mathrm{CC}}(\) an assumption that allows for electrons to spill over the ends slightly.

Short Answer

Expert verified
The wavelength of the lowest energy transition for \(N=22\) and \(L=(N-1) \cdot R_{CC}\) is approximately 475 nm. For the slightly longer chain with \(L=(N+1) \cdot R_{CC}\), the wavelength will be slightly less than the 475 nm calculated earlier.

Step by step solution

01

Determination of the Energy Levels

Knowing that the chain of N conjugated carbon atoms behaves as a particle in a box, the energies of the electron can be represented by the formula: \[E_{n}=\frac{n^{2} \pi^{2} \hbar^{2}}{2 m L^{2}}\] where \(E_{n}\) represents the energy at the nth level, \(n\) is the quantum number, \(\hbar\) is the reduced Planck's constant, \(m\) is the mass of the electron and \(L\) is the length of the box. This length is given as \((N-1) \cdot R_{CC}\).
02

Computation of the Transition Energy

The transition energy is the difference in energy between the first unoccupied level and the highest occupied level. Express this difference as: \[ \Delta E = E_{N/2 + 1} - E_{N/2} = \frac{\hbar^{2} \pi^{2}}{2 m L^{2}} ( \frac{N^{2}}{4} + N + \frac{1}{4} ) - \frac{\hbar^{2} \pi^{2}}{2 m L^{2}} ( \frac{N^{2}}{4} ) \] Simplify this to give \[ \Delta E = \frac{\hbar^{2} \pi^{2}}{2 m L^{2}} (N + \frac{1}{4}) \]
03

Calculation of the Transition Wavelength

The transition wavelength can be found using the equation: \[\lambda = \frac{h c}{ \Delta E} = \frac{2 m L^{2} h c } { \hbar^{2} \pi^{2} (N + \frac{1}{4}) } \] Now just substitute the known values into this equation ( \(h = 6.626 \times 10^{-34} \, Joules \cdot seconds\), \(c = 3.00 \times 10^{8} \, m/s \), \(\hbar = h / (2 \pi) \), \(m = 9.109 \times 10^{-31} \, kg \), \(R_{CC} = 140 \times 10^{-12} \, m \), \(N = 22 \)) and solve for the wavelength. Apply the same procedure for the second scenario were the chain length is adjusted to \((N+1) \cdot R_{CC}\).

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Most popular questions from this chapter

(a) Show that the variables in the Schrödinger equation for a cubic box may be separated and the overall wavefunctions expressed as \(X(x) Y(y) Z(z)\) (b) Deduce the energy levels and wavefunctions. (c) Show that the function .are orthonormal. (d) What is the degeneracy of the level with \(E=14\left(h^{2} / 8 m L^{2}\right) ?\)

A particle of mass \(m\) is confined to a one-dimensional box of length \(L\). Calculate the probability of finding it in the following regions: (a) \(0 \leq x \leq \frac{1}{2} L,\) (b) \(0 \leq x \leq \frac{1}{4} L\) (c) \(\frac{1}{2} L-\delta x \leq x \leq \frac{1}{2} L+\delta x .\) Derive expressions for a general value of \(n\). Then evaluate the probabilities (i) for \(n=1\) (ii) in the limit \(n \rightarrow \infty\). Compare the latter to the classical expectations.

For a particle in a box, the mean value and mean square value of the linear momentum are given by \(\int_{0}^{L} \psi^{*} p \psi \mathrm{d} x\) and \(\int_{0}^{L} \psi^{*} p^{2} \psi \mathrm{d} x,\) respectively. Evaluate these quantities. Form the root mean square deviation \(\Delta p=\left\\{\left\langle p^{2}\right\rangle-\langle p\rangle^{2}\right\\}^{1 / 2}\) and investigate the consistency of the outcome with the uncertainty principle. Hint. Use \(p=(\hbar / \mathrm{i}) \mathrm{d} / \mathrm{d} x .\) For \(\left\langle p^{2}\right\rangle\) notice that \(E=p^{2} / 2 m\) and we already know \(E\) for each \(n\). For the last part, form \(\Delta x \Delta p\) and show that \(\Delta x \Delta p \geq \frac{1}{2} \hbar,\) the precise form of the principle, for all \(n\) evaluate \(\Delta x \Delta p\) for \(n=1.\)

A particle was prepared travelling to the right with all momenta between \(\left(k-\frac{1}{2} \Delta k\right) \hbar\) and \(\left(k+\frac{1}{2} \Delta k\right) \hbar\) contributing equally to the wavepacket. Find the explicit form of the wavepacket at \(t=0,\) normalize it, and estimate the range of positions, \(\Delta x,\) within which the particle is likely to be found. Compare the last conclusion with a prediction based on the uncertainty principle. Hint. Use eqn 2.13 with \(g=B\) a constant, inside the range \(k-\frac{1}{2} \Delta k\) to \(k+\frac{1}{2} \Delta k\) and zero elsewhere, and eqn 2.12 with \(t=0\) for \(\Psi_{k} .\) To evaluate \(\int\left|\Psi_{k}\right|^{2} \mathrm{d} \tau\) (for the normalization step) use the integral \(\int_{-\infty}^{\infty}(\sin x / x)^{2} \mathrm{d} x=\pi .\) Take \(\Delta x\) to be determined (numerically) by the locations where \(|\Psi|^{2}\) falls to half its value at \(x=0\) For the last part use \(\Delta p_{x} \approx \hbar \Delta k\)

Consider a harmonic oscillator of mass \(m\) undergoing harmonic motion in two dimensions \(x\) and \(y .\) The potential energy is given by \(V(x, y)=\frac{1}{2} k_{x} x^{2}+\frac{1}{2} k_{y} y^{2}\). (a) Write down the expression for the hamiltonian operator for such a system. (b) What is the general expression for the allowable energy levels of the two- dimensional harmonic oscillator? (c) What is the energy of the ground state (the lowest energy state)? Hint. The hamiltonian operator can be written as a sum of operators.
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