CHEM106: Assignment 5 Particle in a Box 1. For a one-dimensional particle in a box system, the solution of the Schrödinger n 2h 2 equation leads to the quantized energy E n . What is the zero-point energy for 8mL2 the system? The zero-point energy (ZPE) for a quantum mechanical system is defined as the lowest possible energy allowed. In the case of the harmonic oscillator, the energy depends on the quantum number n, and n = 1, 2, 3, …. 12 h 2 h2 Thus, . ZPE E1 8mL2 8mL2 2. The -electron of a conjugated polyene molecule can be modeled as a particle in a box. Assuming a length of 0.6 nm for the one-dimensional box, A. calculate the energy gap between the first excited state and the ground state for the electron. For the one-dimensional particle in a box, the energy is given by n 2h 2 . En 8mL2 The energy gap between the first excited state (n = 2) and the ground state (n = 1) is DE = E2 - E1 = 22 h2 12 h 2 8me L2 8me L2 = 3h 2 8me L2 3´ (6.626 ´10 -34 Js)2 = 8 ´ 9.1093897 ´10 -31 kg ´ (6 ´10 -10 m)2 = 5.02 ´10 -19 J. B. what is the corresponding frequency of the photon absorbed as the electron makes a transition from the ground state to the first excited state? E h E / h 5.02 10 19 J / 6.626 10 34 Js 7.58 1014 Hz 3. For a one-dimensional particle in a box system, the solution of the Schrödinger 2 n equation generates a wave function n ( x ) sin( x ) for the state n. Verify that L L n (x ) satisfies the Schrödinger equation. For the one-dimensional particle in a box system, the Hamiltonian operator can be defined as 2 d 2 2 d 2 ˆ ˆ ˆ . H T V 0 2m dx 2 2m dx 2 When we operate the Hamiltonian on the wave function, we obtain 2 d2 2 np Ĥ Y n (x) = sin( x) 2 2m dx L L = np 2 2 np ) sin( x) 2m L L L 2 ( n2h2 2 np sin( x) 2 8mL L L = En Y n (x). Thus, the given wave function n (x ) satisfies the Schrödinger equation. = 4. For the ground state of the particle in a box system 1 ( x ) 2 x , calculate the sin L L probability of finding the particle between x = 0 and x = L/2. [Useful integral: sin 2 (x )d x 1 1 x sin( 2x ) ] 2 4 The probability of finding the particle between x = 0 and x = L/2 is L/l L/2 L/2 2 p * ( x )( x )d x sin( x ) sin( x )d x sin 2 ( x )d x . L 0 L L L 0 0 According to the given above, sin 2 (x )d x we have 1 1 x sin( 2x ) , 2 4 2 1 L 2p L/2 [ xsin( x)] 0 L 2 4p L 2 L L 2p L = { [sin( ) - sin(0)]} L 4 4p L 2 2L 1 = = . L4 2 P= 5. Find out the degeneracies of the lowest four energy levels for the three- dimensional particle in a box system with the dimension Lx = Ly = Lz. The energy for the three-dimensional particle in a box system is given by E Ex E y Ez n x2 h 2 8mL2x n 2y h 2 8mL2y n z2 h 2 8mL2z . If Lx = Ly = Lz, the above expression reduces to h2 E (n x2 n 2y n z2 ) . 2 8mLx The degeneracy of an energy level equals to the number of quantum states for the given energy. The lowest four energy levels and the corresponding degeneracy is listed below: Energy Levels h2 (in units of ) 8mL2x 3 6 9 12 Possible States nx ny nz Degeneracy 111 1 2 1, 2 1 1, 1 1 2 2 1 2, 2 2 1, 1 2 2 222 1 3 3 1