Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 39 - Wave Functions and Uncertainty - Exercises and Problems - Page 1175: 17

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ We know that the area under the probability density curve is 1. So we have to draw the $|\psi(x)|^2$ graph to find $c$. We know that the $|\psi(x)|^2$ graph is the square of $|\psi(x)|$. This means that the maximum point on the $|\psi(x)|$-graph, which is $c$, will be $c^2$, and the minimum point of $-c$ will be $ c^2$ as well. The graph is shown below. Now we can find $c$; where the area under this curve consists of one big rectangle. $$A=(8)(c^2) =1$$ Hence, $$c=\sqrt{\dfrac{1}{8}}=\color{red}{\bf 0.354}\;\rm nm^{-1/2}$$ $$\color{blue}{\bf [b]}$$ The graph is shown below. $$\color{blue}{\bf [c]}$$ We know that the probability of detecting a neutron is given by $$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x =\text{Area under the curve}$$ So, $$\text{Prob}( \text{in } -1\leq x\leq 1 ) =\text{Area under the $|\psi(x)|^2$ curve}$$ So that $$\text{Prob}( \text{in } -1\leq x\leq 1 ) =(2)(c^2)=(2)\left( \frac{1}{8}\right)$$ $$\text{Prob}( \text{in } -1\leq x\leq 1 ) = \color{red}{\bf 0.25}$$
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