Question

Given the wave function of a particle \( \psi(x) = \sqrt{\frac{2}{L}} \sin \left( \frac{\pi x}{L} \right) \) for \( 0<x<L \) and 0 elsewhere, the probability of finding the particle between \( x = 0 \) and \( x = L/2 \) is .............

Hint:

The probability of finding a particle in a given region is the integral of \( |\psi(x)|^2 \) over that region. Use trigonometric identities to simplify the integrals when necessary.

Answer 1

Correct Answer: 0.5

Step 1: Probability of finding the particle.
The probability \( P \) of finding the particle between \( x = 0 \) and \( x = L/2 \) is given by the integral of \( |\psi(x)|^2 \) over that range: \[ P = \int_0^{L/2} |\psi(x)|^2 \, dx \]
Step 2: Square the wave function.
The wave function is given by: \[ \psi(x) = \sqrt{\frac{2}{L}} \sin \left( \frac{\pi x}{L} \right) \] Thus: \[ |\psi(x)|^2 = \left( \sqrt{\frac{2}{L}} \sin \left( \frac{\pi x}{L} \right) \right)^2 = \frac{2}{L} \sin^2 \left( \frac{\pi x}{L} \right) \]
Step 3: Set up the integral.
The probability is: \[ P = \int_0^{L/2} \frac{2}{L} \sin^2 \left( \frac{\pi x}{L} \right) \, dx \]
Step 4: Simplify the integral.
Use the identity \( \sin^2 \theta = \frac{1 - \cos(2\theta)}{2} \) to simplify the integral: \[ P = \frac{2}{L} \int_0^{L/2} \frac{1 - \cos \left( \frac{2\pi x}{L} \right)}{2} \, dx \] This simplifies to: \[ P = \frac{1}{L} \int_0^{L/2} \left( 1 - \cos \left( \frac{2\pi x}{L} \right) \right) \, dx \] Evaluating the integral, we get: \[ P = 0.5 \]
Step 5: Conclusion.
Thus, the probability of finding the particle between \( x = 0 \) and \( x = L/2 \) is 0.5.