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Chapter 7: Problem 82

Find the expectation value of the position squared when the particle in the box is in its third excited state and the length of the box is \(L\)

Short Answer

Expert verified
The expectation value of the position squared (\(⟨x^2⟩\)) for a particle in a box in its third excited state, with a length of L, is \(\frac{5L^2}{8}\).

Step by step solution

01

Determine the wavefunction for the third excited state

For the third excited state, n = 4. Substitute this into the general wavefunction formula for a particle in a box: \[ ψ_4(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{4πx}{L}\right) \]
02

Substitute the wavefunction and its complex conjugate into the expectation value formula

In the context of the particle in a box problem, the wavefunction ψ(x) is real, so the complex conjugate is simply equal to the wavefunction itself. Substitute ψ(x) and ψ^*(x) into the expectation value formula: \[ ⟨x^2⟩ = \int_{0}^{L}\left(\sqrt{\frac{2}{L}}\sin\left(\frac{4πx}{L}\right)\right)^2 x^2 dx \] Simplify the expression inside the integral: \[ ⟨x^2⟩ = \int_{0}^{L}\frac{2x^2}{L}\sin^2\left(\frac{4πx}{L}\right) dx \]
03

Evaluate the integral

Now, we need to evaluate the integral to find the expectation value of the position squared: \[ ⟨x^2⟩ = \frac{2}{L}\int_{0}^{L}x^2\sin^2\left(\frac{4πx}{L}\right) dx \] Using integration by parts (or a reference table for integrals), we can find the following result for the integral: \[ ⟨x^2⟩ = \frac{2}{L}\left(\frac{L^3}{12} - \frac{L^3}{32}\right) = \frac{5L^2}{8} \] Thus, the expectation value of the position squared for a particle in a box in its third excited state, with a length of L, is \(\frac{5L^2}{8}\).

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Most popular questions from this chapter

Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1 -eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon's average frequency is this?

A free proton has a wave function given by \(\Psi(x, t)=A e^{i\left(5.02 \times 10^{11} x-8.00 \times 10^{15} t\right)}\) The coefficient of \(x\) is inverse meters \(\left(\mathrm{m}^{-1}\right)\) and the coefficient on \(t\) is inverse seconds \(\left(\mathrm{s}^{-1}\right) .\) Find its momentum and energy.

Is it possible to measure energy of \(0.75 \hbar \omega\) for a quantum harmonic oscillator? Why? Why not? Explain.

Determine the expectation value of the potential energy for a quantum harmonic oscillator in the ground state. Use this to calculate the expectation value of the kinetic energy.

An electron with kinetic energy \(2.0 \mathrm{MeV}\) encounters a potential energy barrier of height \(16.0 \mathrm{MeV}\) and width 2.00 nm. What is the probability that the electron emerges on the other side of the barrier?
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