7. One electron is trapped in a one-dimensional square well potential with infinitely high sides. a. If you have a probe that has a width for electron detection Ax = 0.00350L in the x direction, for the first excited state ( n =2), what is the probability that the electron is found in the probe when it is centered at x = L/4,(hint: you can use an approximation for this - you do not need to do an integral)? b. What is the average number of electrons that you would detect using the probe described in part "b." centered at x= L/4, if the electron is in the first excited state (n = 2 ) for each experiment and you repeat the experiment N, = 100,000 times?

icon
Related questions
Question

Please include significant figures and units. Thanks for your help!

**Quantum Mechanics Problem: Electron in a One-Dimensional Square Well**

**Problem Statement:**
One electron is trapped in a one-dimensional square well potential with infinitely high sides.

**Questions:**
a. If you have a probe that has a width for electron detection \(\Delta x = 0.00350L\) in the x direction, for the first excited state \((n = 2)\), what is the probability that the electron is found in the probe when it is centered at \(x = L/4\)? (Hint: you can use an approximation for this - you do not need to do an integral).

b. What is the average number of electrons that you would detect using the probe described in part "a." centered at \(x = L/4\), if the electron is in the first excited state \((n = 2)\) for each experiment and you repeat the experiment \(N_p = 100,000\) times?

**Solution Explanation:**

**Part (a):**
To determine the probability that the electron is found in the probe when it is centered at \(x = L/4\), we need to evaluate the wave function of the electron in the first excited state \((n = 2)\) at \(x = L/4\). The wave function for a particle in a one-dimensional infinite potential well is given by:
\[
\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)
\]
For the first excited state \((n = 2)\), the wave function becomes:
\[
\Psi_2(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right)
\]
To find the probability of locating the electron within the small region \(\Delta x\) around \(x = L/4\), we use the probability density \(|\Psi_2(x)|^2\):
\[
\text{Probability density} = |\Psi_2(x)|^2 = \left(\sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right)\right)^2 = \frac{2}{L} \sin^2\left(\frac{2\pi x}{L}\right)
\]
Substituting \(x = L/
Transcribed Image Text:**Quantum Mechanics Problem: Electron in a One-Dimensional Square Well** **Problem Statement:** One electron is trapped in a one-dimensional square well potential with infinitely high sides. **Questions:** a. If you have a probe that has a width for electron detection \(\Delta x = 0.00350L\) in the x direction, for the first excited state \((n = 2)\), what is the probability that the electron is found in the probe when it is centered at \(x = L/4\)? (Hint: you can use an approximation for this - you do not need to do an integral). b. What is the average number of electrons that you would detect using the probe described in part "a." centered at \(x = L/4\), if the electron is in the first excited state \((n = 2)\) for each experiment and you repeat the experiment \(N_p = 100,000\) times? **Solution Explanation:** **Part (a):** To determine the probability that the electron is found in the probe when it is centered at \(x = L/4\), we need to evaluate the wave function of the electron in the first excited state \((n = 2)\) at \(x = L/4\). The wave function for a particle in a one-dimensional infinite potential well is given by: \[ \Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \] For the first excited state \((n = 2)\), the wave function becomes: \[ \Psi_2(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right) \] To find the probability of locating the electron within the small region \(\Delta x\) around \(x = L/4\), we use the probability density \(|\Psi_2(x)|^2\): \[ \text{Probability density} = |\Psi_2(x)|^2 = \left(\sqrt{\frac{2}{L}} \sin\left(\frac{2\pi x}{L}\right)\right)^2 = \frac{2}{L} \sin^2\left(\frac{2\pi x}{L}\right) \] Substituting \(x = L/
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer