Chapter 34, Problem 49P

(a)

To determine

The probability that the particle will be found in the region.

Answer to Problem 49P

The probability that the particle will be found in the region is $0.50$ .

Explanation of Solution

Given:

The one-dimensional box region is $0\le x\le L$ .

The particle is in the first excited state.

The given region is $0<x<\frac{L}{2}$ .

Formula used:

The expression for probability for finding the particle in first excited state is given by,

  $P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx$

From integral formula,

  ${\displaystyle \int {\sin }^{2}d\theta =}\frac{\theta }{2}-\frac{\sin 2\theta }{4}+c$

Calculation:

Let, $\frac{2\pi x}{L}=\theta$ .

By differentiating both sides,

  $\begin{array}{c}\frac{2\pi dx}{L}=d\theta \\ dx=\frac{Ld\theta }{2\pi }\end{array}$

The limit is $0\text{to}\frac{L}{2}$ changes to $0\text{to}\pi$ .

The probability is calculated as,

  $\begin{array}{c}P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{L/2}{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{2}{L}{\sin }^2}\theta \left(\frac{Ld\theta }{2\pi }\right)\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^2}\theta d\theta \end{array}$

Solving further as,

  $\begin{array}{c}P=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{\sin }^2}\theta d\theta \\ =\frac{1}{\pi }{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right]}_0^{\pi }\\ =\frac{1}{\pi }\left[\frac{\pi }{2}-0\right]\\ =0.5\end{array}$

Conclusion:

Therefore, the probability that the particle will be found in the region is $0.50$ .

(b)

To determine

The probability that the particle will be found in the region.

Answer to Problem 49P

The probability that the particle will be found in the region is $0.402$ .

Explanation of Solution

Given:

The given region is $0<x<\frac{L}{3}$ .

Formula used:

The expression for probability for finding the particle in first excited state is given by,

  $P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx$

From integral formula,

  ${\displaystyle \int {\sin }^{2}d\theta =}\frac{\theta }{2}-\frac{\sin 2\theta }{4}+c$

Calculation:

Let, $\frac{2\pi x}{L}=\theta$ .

By differentiating both sides,

  $\begin{array}{c}\frac{2\pi dx}{L}=d\theta \\ dx=\frac{Ld\theta }{2\pi }\end{array}$

The limit is $0\text{to}\frac{L}{3}$ changes to $0\text{to}\frac{2\pi }{3}$ .

The probability is calculated as,

  $\begin{array}{c}P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{L/3}{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{2\pi /3}{\int }}\frac{2}{L}{\sin }^2}\theta \left(\frac{Ld\theta }{2\pi }\right)\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi /3}{\int }}{\sin }^2}\theta d\theta \end{array}$

Solving further as,

  $\begin{array}{c}P=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi /3}{\int }}{\sin }^2}\theta d\theta \\ =\frac{1}{\pi }{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right]}_0^{2\pi /3}\\ =\frac{1}{\pi }\left[\frac{\pi }{3}-\frac{\sin 4\pi /3}{4}\right]\\ =0.402\end{array}$

Conclusion:

Therefore, the probability that the particle will be found in the region is $0.402$ .

(c)

To determine

The probability that the particle will be found in the region.

Answer to Problem 49P

The probability that the particle will be found in the region is $0.750$ .

Explanation of Solution

Given:

The given region is $0<x<\frac{3L}{4}$ .

Formula used:

The expression for probability for finding the particle in ground state is given by,

  $P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx$

From integral formula,

  ${\displaystyle \int {\sin }^{2}d\theta =}\frac{\theta }{2}-\frac{\sin 2\theta }{4}+c$

Calculation:

Let, $\frac{2\pi x}{L}=\theta$ .

By differentiating both sides,

  $\begin{array}{c}\frac{2\pi dx}{L}=d\theta \\ dx=\frac{Ld\theta }{2\pi }\end{array}$

The limit is $0\text{to}\frac{3L}{4}$ changes to $0\text{to}\frac{3\pi }{2}$ .

The probability is calculated as,

  $\begin{array}{c}P={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{3L/4}{\int }}\frac{2}{L}{\sin }^2\frac{2\pi x}{L}}dx\\ ={\displaystyle \underset{0}{\overset{3\pi /2}{\int }}\frac{2}{L}{\sin }^2}\theta \left(\frac{Ld\theta }{2\pi }\right)\\ =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{3\pi /2}{\int }}{\sin }^2}\theta d\theta \end{array}$

Solving further as,

  $\begin{array}{c}P=\frac{1}{\pi }{\displaystyle \underset{0}{\overset{3\pi /2}{\int }}{\sin }^2}\theta d\theta \\ =\frac{1}{\pi }{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right]}_0^{3\pi /2}\\ =\frac{1}{\pi }\left[\frac{3\pi }{4}-0\right]\\ =0.750\end{array}$

Conclusion:

Therefore, the probability that the particle will be found in the region is $0.750$ .