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Chapter 7: Problem 94

94\. A particle of mass \(m\) confined to a box of width \(L\) is in its first excited state \(\psi_{2}(x) .\) (a) Find its average position (which is the expectation value of the position). (b) Where is the particle most likely to be found?

Short Answer

Expert verified
The average position or expectation value of the particle is \(L/2\). The particle is most likely to be found at positions \(L/4\) and \(3L/4\).

Step by step solution

01

State the Wavefunction

The wave function for a particle in a 1-dimensional box in its first excited state is given by \(\psi_{2}(x) = \sqrt{2/L} \sin(2\pi x /L)\). The limits of the box are 0 and L.
02

Determine the Average Position

The average or expectation value of the position is given by \(\langle x \rangle = \int_0^L x |\psi_{2}(x)|^2 dx\). We square the wave function to get the probability density. Simplify this integral and compute its value.
03

Find Where the Particle is Most Likely Found

The particle is most likely found at the position where the squared wave function, or the probability density \(|\psi_{2}(x)|^2\), is maximized. Set the derivative of the probability density function equal to zero and solve for x to find the maximum.

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