Consider a particle of mass m confined to a one-dimensional box of length L and in a state with normalized wavefunction ψn. (a) Without evaluating any integrals, explain why ⟨x⟩ = L/2. (b) Without evaluating any integrals, explain why ⟨px⟩ = 0. (c) Derive an expression for ⟨x2⟩ (the necessary integrals will be found in the Resource section). (d) For a particle in a box the energy is given by En = n2h2/8mL2 and, because the potential energy is zero, all of this energy is kinetic. Use this observation and, without evaluating any integrals, explain why <p2x> = n2h2/4L2.
Consider a particle of mass m confined to a one-dimensional box of length L and in a state with normalized wavefunction ψn. (a) Without evaluating any integrals, explain why ⟨x⟩ = L/2. (b) Without evaluating any integrals, explain why ⟨px⟩ = 0. (c) Derive an expression for ⟨x2⟩ (the necessary integrals will be found in the Resource section). (d) For a particle in a box the energy is given by En = n2h2/8mL2 and, because the potential energy is zero, all of this energy is kinetic. Use this observation and, without evaluating any integrals, explain why <p2x> = n2h2/4L2.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Consider a particle of mass m confined to a one-dimensional box of length L and in a state with normalized wavefunction ψn. (a) Without evaluating any integrals, explain why ⟨x⟩ = L/2. (b) Without evaluating any integrals, explain why ⟨px⟩ = 0. (c) Derive an expression for ⟨x2⟩ (the necessary integrals will be found in the Resource section). (d) For a particle in a box the energy is given by En = n2h2/8mL2 and, because the potential energy is zero, all of this energy is kinetic. Use this observation and, without evaluating any integrals, explain why <p2x> = n2h2/4L2.
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