Question

An electron is trapped in a one-dimensional infinite potential well that is \(L=300 .\) pm wide. What is the probability that the electron in the first excited state will be detected in an interval between \(x=0.500 L\) and \(x=0.750 L ?\)

Short answer

Answer: There is a 25% chance of detecting the electron in the first excited state between x=0.5L and x=0.75L in a one-dimensional infinite potential well with a width of 300 pm.

Step-by-step solution

Finding the wave function for the first excited state

The wave function for a particle in a one-dimensional infinite potential well of width L can be described by the following equation: \(\psi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)\), where n is the energy level of the particle (1 for the ground state, 2 for the first excited state, and so on). Since we are asked to find the probability for the electron in the first excited state, we should use n=2 to find the wave function for the electron: \(\psi_2(x) = \sqrt{\frac{2}{300\,\text{pm}}}\sin\left(\frac{2\pi x}{300\,\text{pm}}\right)\).

Finding the probability density function

To find the probability of detecting the electron, we need to square the wave function, which will give us the probability density function: \(|\psi_2(x)|^2 = \left(\sqrt{\frac{2}{300\,\text{pm}}}\sin\left(\frac{2\pi x}{300\,\text{pm}}\right)\right)^2 = \frac{2}{300\,\text{pm}}\sin^2\left(\frac{2\pi x}{300\,\text{pm}}\right)\).

Integrating the probability density function over the given interval

Now we need to integrate the probability density function over the interval between x=0.5L and x=0.75L to find the probability of detecting the electron in this region: \(\int_{0.5(300\,\text{pm})}^{0.75(300\,\text{pm})}\frac{2}{300\,\text{pm}}\sin^2\left(\frac{2\pi x}{300\,\text{pm}}\right)dx\).

Evaluating the integral

To evaluate the integral, we first convert the sin^2 function to an equivalent expression using the double angle formula: \(\sin^2(x) = \frac{1-\cos(2x)}{2}\). This allows us to rewrite the integral as: \(\int_{0.5(300\,\text{pm})}^{0.75(300\,\text{pm})}\frac{2}{300\,\text{pm}}\frac{1-\cos\left(\frac{4\pi x}{300\,\text{pm}}\right)}{2}dx\). Now, we can integrate the two parts separately: \( \frac{1}{300\,\text{pm}}\int_{0.5(300\,\text{pm})}^{0.75(300\,\text{pm})}(1-\cos\left(\frac{4\pi x}{300\,\text{pm}}\right))dx\). After evaluating and combining the two integrals, we find that the final probability is approximately: \(P = 0.25\). Thus, there is a 25% chance of detecting the electron in the first excited state between x=0.5L and x=0.75L in a one-dimensional infinite potential well with a width of 300 pm.