Page 112 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:36      9in x 6in     b985-ch09     Elementary Physical Chemistry















                                                      Chapter 9

                                     Applications of Quantum Theory







                            9.1. Translational Motion. Particle-in-a-Box

                            Consider a particle of mass m in a one-dimensional box of length L.
                            Assuming that the potential energy is V = 0 inside box and V = ∞ outside
                                                         o
                            the box, the solution of the Schr¨dinger equation yields the energy (see
                            Fig. 9.1):
                                                    2
                                                       2
                                              E n = n h /8 mL 2  n =1, 2,...            (9.1)
                            and wave-function
                                           ψ n =(2/L) 1/2  sin(nπx/L) n =1, 2,...       (9.2)

                            Note that the spacing between energy levels increases with increasing n and
                            decreases with increasing L.When L becomes very large (of macroscopic
                            dimensions), the energy distribution becomes practically continuous.
                            Example 9.1. A conjugated polyene molecule is sometimes simulated by
                            a one-dimensional-box. If L =2.0 nm and an electron in the box is excited
                            from state 5 to state 6, what is the transition energy?

                            Solution
                            The energy difference between level 5 and level 6 is
                                                         2
                                                  2
                                     ∆E 6←5 =11 h /(8 mL )
                                                                       ) J s
                                                     11 × (6.6260 × 10 −34 2  2 2
                                            =
                                                                              ) m
                                              8 × 9.10939 × 10 −31  kg × (2.0 × 10 −9 2  2
                                            =1.6554 × 10 −19  J ≈ 1.0 eV                (9.3)
                                                                   2 −2
                               [Note: I J =1 Nm=1kg ms −2  m= 1kgm s   ;1 eV = 1.607 × 10 −19  J]
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