Question

In the case of a particle in a one-dimensional infinite potential well (box), what is the probability of finding the particle in the first half of the box for the ground state?

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{4} \)
• D. \( 1 \)

Hint:

In quantum mechanics, the probability of finding a particle in a given region is proportional to the square of the wave function. For the ground state in a box, the probability is evenly distributed over the entire box for simple problems like this.

Answer 1

The Correct Option is A

Step 1: Understanding the problem. In the case of a particle confined in a one-dimensional infinite potential well (also known as a particle in a box), the probability of finding the particle at any point within the box is proportional to the square of the wave function \( \psi(x) \). For the ground state, the wave function is given by: \[ \psi(x) = \sqrt{\frac{2}{L}} \sin \left( \frac{\pi x}{L} \right) \] where: - \( L \) is the length of the box, - \( x \) is the position within the box (ranging from 0 to \( L \)). Step 2: Calculating the probability. The probability of finding the particle in a specific region is given by the square of the wave function integrated over that region. The probability of finding the particle in the first half of the box is: \[ P_{\text{first half}} = \int_0^{L/2} \left| \psi(x) \right|^2 dx \] Substitute the expression for \( \psi(x) \): \[ P_{\text{first half}} = \int_0^{L/2} \left( \frac{2}{L} \sin^2 \left( \frac{\pi x}{L} \right) \right) dx \] This integral can be solved, and the result is: \[ P_{\text{first half}} = \frac{1}{2} \] Step 3: Conclusion. Thus, the probability of finding the particle in the first half of the box for the ground state is \( \frac{1}{2} \). Answer: Therefore, the probability of finding the particle in the first half of the box is \( \frac{1}{2} \).