Question

Q. A particle is confined between rigid walls by a distance L.
(a) Show that the probability P that it will be found within a distance L/3 from one wall is given by
P=1/3[1-sin⁡〖2nπ/3〗/(2nπ/3)]
(b) Evaluate probability for (i) n=1, (ii) n=2, (iii) n=3

Answer 1

Answer on Question #59929-Physics-Quantum Mechanics

Q. A particle is confined between rigid walls by a distance $L$.

(a) Show that the probability $P$ that it will be found within a distance $L/3$ from one wall is given by

$$P = \frac{1}{3} \left[ 1 - \frac{\sin \frac{2n\pi}{3}}{\frac{2n\pi}{3}} \right]$$

(b) Evaluate probability for (i) $n=1$, (ii) $n=2$, (iii) $n=3$

Solution

(a) A wave function is

$$\psi_n(x) = \sqrt{\frac{2}{L}} \sin \frac{n\pi x}{L}$$

The probability $P$ that it will be found within a distance $L/3$ from one wall is

$$P = \int_{0}^{\frac{L}{3}} |\psi_n|^2 dx = \int_{0}^{\frac{L}{3}} \frac{2}{L} \sin^2 \frac{n\pi x}{L} dx = \frac{2}{L} \left( \int_{0}^{\frac{L}{3}} \left( \frac{1}{2} - \frac{1}{2} \cos \frac{2n\pi x}{L} \right) dx \right).$$ $$P = \left( \frac{x}{L} - \frac{\sin \frac{2n\pi x}{L}}{2n\pi} \right)_{0}^{\frac{L}{3}} = \frac{1}{3} \left[ 1 - \frac{\sin \frac{2n\pi}{3}}{\frac{2n\pi}{3}} \right]$$

(b)(i)

$$P(n=1) = \frac{1}{3} \left[ 1 - \frac{\sin \frac{2\pi}{3}}{\frac{2\pi}{3}} \right] = \frac{1}{3} \left[ 1 - \frac{\sqrt{3}}{\frac{2}{\pi}} \right] = \frac{1}{3} \left[ 1 - \frac{3\sqrt{3}}{4\pi} \right] = 0.196$$

(ii)

$$P(n=2) = \frac{1}{3} \left[ 1 - \frac{\sin \frac{4\pi}{3}}{\frac{4\pi}{3}} \right] = \frac{1}{3} \left[ 1 + \frac{\sqrt{3}}{\frac{4\pi}{3}} \right] = \frac{1}{3} \left[ 1 + \frac{3\sqrt{3}}{8\pi} \right] = 0.402$$

(iii)

$$P(n=3) = \frac{1}{3} \left[ 1 - \frac{\sin 2\pi}{2\pi} \right] = \frac{1}{3} [1 - 0] = \frac{1}{3} = 0.333$$

https://www.AssignmentExpert.com