Chapter 34, Problem 48P

(a)

To determine

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

Answer to Problem 48P

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

Explanation of Solution

Given:

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

The particle is in the ground state.

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

Formula used:

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

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

From integral formula,

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

Calculation:

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

By differentiating both sides,

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

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

The probabilityis calculated as,

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

Solving further as,

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

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 48P

The probability that the particle will be found in the region is $0.196$ .

Explanation of Solution

Given:

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

Formula used:

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

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

From integral formula,

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

Calculation:

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

By differentiating both sides,

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

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

The probability is calculated as,

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

Solving further as,

  $\begin{array}{c}P=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\pi /3}{\int }}{\sin }^2}\theta d\theta \\ =\frac{2}{\pi }{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right]}_0^{\pi /3}\\ =\frac{2}{\pi }\left[\frac{\pi }{6}-\frac{\sin 2\pi /3}{4}\right]\\ =0.196\end{array}$

Conclusion:

Therefore, the probability that the particle will be found in the region is $0.196$ .

(c)

To determine

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

Answer to Problem 48P

The probability that the particle will be found in the region is $0.909$ .

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,

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

From integral formula,

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

Calculation:

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

By differentiating both sides,

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

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

The probability is calculated as,

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

Solving further as,

  $\begin{array}{c}P=\frac{2}{\pi }{\displaystyle \underset{0}{\overset{3\pi /4}{\int }}{\sin }^2}\theta d\theta \\ =\frac{2}{\pi }{\left[\frac{\theta }{2}-\frac{\sin 2\theta }{4}\right]}_0^{3\pi /4}\\ =\frac{2}{\pi }\left[\frac{3\pi }{8}-\frac{\sin 6\pi /4}{4}\right]\\ =0.909\end{array}$

Conclusion:

Therefore, the probability that the particle will be found in the region is $0.909$ .