Question

A particle is in a square infinite potential well of width \(L\) and is in the \(n=3\) state. What is the probability that, when observed, the particle is found to be in the rightmost \(10.0 \%\) of the well?

Short answer

The probability of finding the particle in the rightmost 10% of the well is 20%.

Step-by-step solution

Determine the wavefunction for the n=3 state

The wavefunction for a particle in an infinite square well is given by: $$ \psi_n(x) = \sqrt{\frac{2}{L}}\sin{\frac{n\pi}{L}x} $$ For the n=3 state, we simply substitute \(n=3\) into the wavefunction expression: $$ \psi_3(x) = \sqrt{\frac{2}{L}}\sin{\frac{3\pi}{L}x} $$ This is the wavefunction we'll use to calculate the probability.

Define the region of interest

We are asked to find the probability of the particle being in the rightmost 10% of the well. In terms of width L, this corresponds to a region from \(x = 0.9L\) to \(x = L\).

Calculate the probability

The probability of finding the particle in a particular region is given by the integral of the square of the wavefunction over that region: $$ P = \int_{x_1}^{x_2} |\psi_3(x)|^2 dx $$ In this case, the limits of integration are \(x_1 = 0.9L\) and \(x_2 = L\). We will now evaluate the integral: $$ P = \int_{0.9L}^{L} |\sqrt{\frac{2}{L}}\sin{\frac{3\pi}{L}x}|^2 dx $$ Simplify the integral: $$ P = \frac{2}{L} \int_{0.9L}^{L} \sin^2{\frac{3\pi}{L}x} dx $$ Now we need to compute this integral. We can use a double angle formula, \(\sin^2(u) = \frac{1}{2}(1-\cos(2u))\), to make the integration process easier. Let us rewrite the integral using this formula: $$ P = \frac{2}{L} \int_{0.9L}^{L} \frac{1}{2}(1-\cos(\frac{6\pi}{L}x)) dx $$ Now, we can integrate: $$ P = \frac{2}{L} \left[ \frac{1}{2}x - \frac{1}{4}\frac{L}{6\pi}\sin(\frac{6\pi}{L}x) \right]_{0.9L}^L $$ Finally, we will substitute the limits of integration, \(0.9L\) and \(L\). $$ P = \frac{2}{L} \left[\frac{1}{2}L - \frac{1}{4}\frac{L}{6\pi}\sin(6\pi) - (\frac{1}{2}(0.9L) - \frac{1}{4}\frac{L}{6\pi}\sin(5.4\pi)) \right] $$ Since \(\sin(6\pi) = 0\) and \(\sin(5.4\pi) = 0\), we can simplify this further: $$ P = \frac{2}{L}\left[\frac{1}{2}L - \frac{1}{2}(0.9L)\right] = \frac{2}{L}(\frac{1}{10}L) $$ The L's cancel out: $$ P = \frac{2}{10} = \boxed{0.2} $$ So, the probability of finding the particle in the rightmost 10% of the well when observed is 20%.