Chapter 28, Problem 55P

(a)

To determine

The energy of the particle in terms of $h$, $m$ and $L$.

Answer to Problem 55P

The energy of the particle in terms of $h$, $m$ and $L$ is $E=\frac{h^2}{4{\pi }^{2}m{L}^2}$.

Explanation of Solution

Write the Schrodinger’s equation.

  $-\frac{{\hslash }^2}{2m}\frac{d^2\Psi }{d{x}^2}+U\Psi =E\Psi$        (I)

Here, $\hslash$ is the reduced Planck’s constant, $m$ is the mass of the particle, $\Psi$ is the given wave function, $U$ is the potential energy and $E$ is the total energy of the particle.

Write the equation for the potential energy.

  $U\left(x\right)=\frac{-{\hslash }^2{x}^2}{m{L}^2\left(L^2-x^2\right)}$        (II)

Here, $L$ is the half of the length of the potential well.

Write the expression of the given wavefunction.

  $\Psi \left(x\right)=A\left(1-\frac{x^2}{L^2}\right)$        (III)

Take the derivative of the above wave function with respect to $x$.

  $\begin{array}{c}\frac{d\Psi }{dx}=A\left(0-\frac{2x}{L^2}\right)\\ =-\frac{2Ax}{L^2}\end{array}$

Take the derivative of the above equation with respect to $x$.

  $\begin{array}{c}\frac{d^2\Psi }{d^{2}x}=\frac{d}{dx}\left(-\frac{2Ax}{L^2}\right)\\ =-\frac{2A}{L^2}\end{array}$        (IV)

Put equations (II), (III) and (IV) in equation (I) and rearrange it.

  $\begin{array}{c}-\frac{{\hslash }^2}{2m}\left(-\frac{2A}{L^2}\right)+\left(\frac{-h^2{x}^2}{m{L}^2\left(L^2-x^2\right)}\right)A\left(1-\frac{x^2}{L^2}\right)=EA\left(1-\frac{x^2}{L^2}\right)\\ \frac{{\hslash }^{2}A}{m{L}^2}-\frac{{\hslash }^2{x}^{2}A}{m{L}^2\left(L^2-x^2\right)}\frac{\left(L^2-x^2\right)}{L^2}=EA\left(1-\frac{x^2}{L^2}\right)\\ \frac{{\hslash }^2}{m{L}^2}-\frac{{\hslash }^2{x}^2}{m{L}^4}=E\left(1-\frac{x^2}{L^2}\right)\\ \frac{{\hslash }^2}{m{L}^2}\left(1-\frac{x^2}{L^2}\right)=E\left(1-\frac{x^2}{L^2}\right)\end{array}$        (V)

Equation (V) will be true for all $x$ if

  $E=\frac{{\hslash }^2}{m{L}^2}$        (VI)

Write the equation for the reduced Planck’s constant.

  $\hslash =\frac{h}{2\pi }$        (VII)

Here, $h$ is the Planck’s constant.

Conclusion:

Put equation (VII) in (VI).

  $\begin{array}{c}E=\frac{{\left(\frac{h}{2\pi }\right)}^2}{m{L}^2}\\ =\frac{h^2}{4{\pi }^{2}m{L}^2}\end{array}$

Therefore, the energy of the particle in terms of $h$, $m$ and $L$ is $E=\frac{h^2}{4{\pi }^{2}m{L}^2}$.

(b)

To determine

The normalization constant $A$.

Answer to Problem 55P

The normalization constant $A$ is $\sqrt{\frac{15}{16L}}$.

Explanation of Solution

The given wavefunction is an even wavefunction.

Write the normalization condition for the given even wave function.

  $2{\displaystyle \underset{0}{\overset{L}{\int }}\left|{\Psi }^2\right|dx}=1$        (VIII)

Put equation (III) in equation (VIII) and rearrange it.

  $\begin{array}{c}2{\displaystyle \underset{0}{\overset{L}{\int }}\left|{\left[A\left(1-\frac{x^2}{L^2}\right)\right]}^2\right|dx}=1\\ 2{\displaystyle \underset{0}{\overset{L}{\int }}{A}^2{\left(1-\frac{x^2}{L^2}\right)}^{2}dx}=1\\ 2A^2{\displaystyle \underset{0}{\overset{L}{\int }}\left(1-\frac{2x^2}{L^2}+\frac{x^4}{L^4}\right)dx}=1\end{array}$

Integrate the above equation.

  $\begin{array}{c}2A^2{\left[x-\frac{2x^3}{3L^2}+\frac{x^5}{5L^4}\right]}_0^L=1\\ 2A^2\left(L-\frac{2L^3}{3L^2}+\frac{L^5}{5L^4}-0\right)=1\\ 2A^2\left(L-\frac{2L}{3}+\frac{L}{5}\right)=1\\ A^2\left(\frac{16L}{15}\right)=1\end{array}$

Rearrange the above equation for $A$.

  $\begin{array}{c}{A}^2=\frac{15}{16L}\\ A=\sqrt{\frac{15}{16L}}\end{array}$

Conclusion:

Therefore, the normalization constant $A$ is $\sqrt{\frac{15}{16L}}$.

(c)

To determine

The probability that the particle is located in the range $-L/3\le x\le L/3$.

Answer to Problem 55P

The probability that the particle is located in the range $-L/3\le x\le L/3$ is $0.580$.

Explanation of Solution

Write the equation for the probability that the particle lies in the range $-L/3\le x\le L/3$ for the given even wavefunction.

  $P=2{\displaystyle \underset{0}{\overset{L/3}{\int }}{\Psi }^{2}dx}$        (IX)

Here, $P$ is the probability.

Put equation (III) in equation (IX) and rearrange it.

  $\begin{array}{c}P=2{\displaystyle \underset{0}{\overset{L/3}{\int }}\left|{\left[A\left(1-\frac{x^2}{L^2}\right)\right]}^2\right|dx}\\ =2{\displaystyle \underset{0}{\overset{L/3}{\int }}{A}^2{\left(1-\frac{x^2}{L^2}\right)}^{2}dx}\\ P=2A^2{\displaystyle \underset{0}{\overset{L/3}{\int }}\left(1-\frac{2x^2}{L^2}+\frac{x^4}{L^4}\right)dx}\end{array}$

Integrate the above equation.

  $\begin{array}{c}P=2A^2{\left[x-\frac{2x^3}{3L^2}+\frac{x^5}{5L^4}\right]}_0^{L/3}\\ =2A^2\left(\frac{L}{3}-\frac{2{\left(L/3\right)}^3}{3L^2}+\frac{{\left(L/3\right)}^5}{5L^4}-0\right)\\ =2A^2\left(\frac{L}{3}-\frac{2L^3}{81L^2}+\frac{L^5}{1215L^4}\right)\\ =2A^2\left(\frac{L}{3}-\frac{2L}{81}+\frac{L}{1215}\right)\end{array}$        (X)

Conclusion:

Substitute $\sqrt{\frac{15}{16L}}$ for $A$ in equation (X) to find $P$.

$\begin{array}{c}P=2{\left(\sqrt{\frac{15}{16L}}\right)}^2\left(\frac{L}{3}-\frac{2L}{81}+\frac{L}{1215}\right)\\ =2\frac{15}{16L}\left(\frac{L}{3}-\frac{2L}{81}+\frac{L}{1215}\right)\\ =\frac{47}{81}\\ =0.580\end{array}$

Therefore, the probability that the particle is located in the range $-L/3\le x\le L/3$ is $0.580$.