Chapter 7, Problem 7B.7P

(a)

Interpretation Introduction

Interpretation:

The probability of finding the particle of wavefunction ${\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$ between $x=4.95\text{nm}$ and $x=5.05\text{nm}$ has to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.7P

The probability that the particle is in between $x=4.95\text{nm}$ and $x=5.05\text{nm}$, is $\underset{\_}{0.0098}$.

Explanation of Solution

The probability that the particle is in between $x=4.95\text{nm}$ and $x=5.05\text{nm}$, is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{4.95}{\overset{5.05}{\int }}{\psi }^{\ast }\psi dx}$                                                                                (1)

Where,

• $\psi$ is the normalized wave function.
• ${\psi }^{\ast }$ is the conjugated complex of the normalized wave function.

The given wavefunction of the particle is shown below.

    $\psi ={\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$

The wave function is real.  So, ${\psi }^{\ast }$ is equal to $\psi$.

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (1).

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{4.95}{\overset{5.05}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right){\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\\ ={\displaystyle \underset{4.95}{\overset{5.05}{\int }}\left(2/L\right){\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{4.95}{\overset{5.05}{\int }}{\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{4.95}{\overset{5.05}{\int }}\frac{1}{2}\left(1-\cos 2\left(\pi x/L\right)\right)dx}\end{array}$

The identity used in the above equation is shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration, by substituting the value of $L=10\text{nm}$ and $\pi =3.14$.

    $\begin{array}{c}\text{Probability}=\frac{1}{2}\left(2/10\right){\displaystyle \underset{4.95}{\overset{5.05}{\int }}\left(1-\cos 2\left(\pi x/10\right)\right)dx}\\ =\frac{1}{10}\times {\left[x-\frac{\sin 2\left(\pi x/10\right)}{2\left(\pi /10\right)}\right]}_{4.95}^{5.05}\\ =\frac{1}{10}\times \left[\left(5.05-\frac{\sin 2\left(5.05\times \pi /10\right)}{2\left(\pi /10\right)}\right)-\left(4.95-\frac{\sin 2\left(4.95\times \pi /10\right)}{2\left(\pi /10\right)}\right)\right]\\ =\frac{1}{10}\times \left[\left(5.05-\frac{\sin \left(1.01\pi \right)}{\left(3.14/5\right)}\right)-\left(4.95-\frac{\sin \left(0.99\pi \right)}{\left(3.14/5\right)}\right)\right]\end{array}$

The value of $\sin \left(1.01\pi \right)$ is $0.055$ and $\sin \left(0.99\pi \right)$ is $0.054$.

On further solving the above equation, by substituting the value of $\sin \left(1.01\pi \right)$ and $\sin \left(0.99\pi \right)$.

    $\begin{array}{c}\text{Probability}=\frac{1}{10}\times \left[\left(5.05-\frac{0.055}{0.628}\right)-\left(4.95-\frac{0.054}{0.628}\right)\right]\\ =\frac{1}{10}\times \left[\left(5.05-0.087\right)-\left(4.95-0.085\right)\right]\\ =\frac{1}{10}\times \left[\left(4.963\right)-\left(4.865\right)\right]\\ =\underset{\_}{0.0098}\end{array}$

Hence, the probability that the particle is in between $x=4.95\text{nm}$ and $x=5.05\text{nm}$, is $\underset{\_}{0.0098}$.

(b)

Interpretation Introduction

Interpretation:

The probability of finding the particle of wavefunction ${\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$ between $x=1.95\text{nm}$ and $x=2.05\text{nm}$ has to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.7P

The probability that the particle is in between $x=1.95\text{nm}$ and $x=2.05\text{nm}$, is $\underset{\_}{0.0098}$.

Explanation of Solution

The probability that the particle is in between $x=1.95\text{nm}$ and $x=2.05\text{nm}$, is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{1.95}{\overset{2.05}{\int }}{\psi }^{\ast }\psi dx}$                                                                                (2)

Where,

• $\psi$ is the normalized wave function.
• ${\psi }^{\ast }$ is the conjugated complex of the normalized wave function.

The given wavefunction of the particle is shown below.

    $\psi ={\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$

The wave function is real.  So, ${\psi }^{\ast }$ is equal to $\psi$.

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (2).

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{1.95}{\overset{2.05}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right){\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\\ ={\displaystyle \underset{1.95}{\overset{2.05}{\int }}\left(2/L\right){\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{1.95}{\overset{2.05}{\int }}{\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{1.95}{\overset{2.05}{\int }}\frac{1}{2}\left(1-\cos 2\left(\pi x/L\right)\right)dx}\end{array}$

The identity used in the above equation is shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration, by substituting the value of $L=10\text{nm}$ and $\pi =3.14$.

    $\begin{array}{c}\text{Probability}=\frac{1}{2}\left(2/10\right){\displaystyle \underset{1.95}{\overset{2.05}{\int }}\left(1-\cos 2\left(\pi x/10\right)\right)dx}\\ =\frac{1}{10}\times {\left[x-\frac{\sin 2\left(\pi x/10\right)}{2\left(\pi /10\right)}\right]}_{1.95}^{2.05}\\ =\frac{1}{10}\times \left[\left(2.05-\frac{\sin 2\left(2.05\times \pi /10\right)}{2\left(\pi /10\right)}\right)-\left(1.95-\frac{\sin 2\left(1.95\times \pi /10\right)}{2\left(\pi /10\right)}\right)\right]\\ =\frac{1}{10}\times \left[\left(2.05-\frac{\sin \left(0.41\pi \right)}{\left(3.14/5\right)}\right)-\left(1.95-\frac{\sin \left(0.39\pi \right)}{\left(3.14/5\right)}\right)\right]\end{array}$

The value of $\sin \left(0.41\pi \right)$ is $0.022$ and $\sin \left(0.39\pi \right)$ is $0.021$.

On further solving the above equation, by substituting the value of $\sin \left(1.01\pi \right)$ and $\sin \left(0.99\pi \right)$.

    $\begin{array}{c}\text{Probability}=\frac{1}{10}\times \left[\left(2.05-\frac{0.022}{0.628}\right)-\left(1.95-\frac{0.021}{0.628}\right)\right]\\ =\frac{1}{10}\times \left[\left(2.05-0.035\right)-\left(1.95-0.033\right)\right]\\ =\frac{1}{10}\times \left[\left(2.015\right)-\left(1.917\right)\right]\\ =\underset{\_}{0.0098}\end{array}$

Hence, the probability that the particle is in between $x=1.95\text{nm}$ and $x=2.05\text{nm}$, is $\underset{\_}{0.0098}$.

(c)

Interpretation Introduction

Interpretation:

The probability of finding the particle of wavefunction ${\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$ between $x=9.90\text{nm}$ and $x=10.00\text{nm}$ has to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.7P

The probability that the particle is in between $x=9.90\text{nm}$ and $x=10.00\text{nm}$, is $\underset{\_}{0.0271}$.

Explanation of Solution

The probability that the particle is in between $x=9.90\text{nm}$ and $x=10.00\text{nm}$, is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{9.90}{\overset{10.00}{\int }}{\psi }^{\ast }\psi dx}$                                                                               (3)

Where,

• $\psi$ is the normalized wave function.
• ${\psi }^{\ast }$ is the conjugated complex of the normalized wave function.

The given wavefunction of the particle is shown below.

    $\psi ={\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$

The wave function is real.  So, ${\psi }^{\ast }$ is equal to $\psi$.

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (3).

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{9.90}{\overset{10.00}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right){\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\\ ={\displaystyle \underset{9.90}{\overset{10.00}{\int }}\left(2/L\right){\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{9.90}{\overset{10.00}{\int }}{\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{9.90}{\overset{10.00}{\int }}\frac{1}{2}\left(1-\cos 2\left(\pi x/L\right)\right)dx}\end{array}$

The identity used in the above equation is shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration, by substituting the value of $L=10\text{nm}$ and $\pi =3.14$.

    $\begin{array}{c}\text{Probability}=\frac{1}{2}\left(2/10\right){\displaystyle \underset{9.90}{\overset{10.00}{\int }}\left(1-\cos 2\left(\pi x/10\right)\right)dx}\\ =\frac{1}{10}\times {\left[x-\frac{\sin 2\left(\pi x/10\right)}{2\left(\pi /10\right)}\right]}_{9.90}^{10.00}\\ =\frac{1}{10}\times \left[\left(10.00-\frac{\sin 2\left(10.00\times \pi /10\right)}{2\left(\pi /10\right)}\right)-\left(9.90-\frac{\sin 2\left(9.90\times \pi /10\right)}{2\left(\pi /10\right)}\right)\right]\\ =\frac{1}{10}\times \left[\left(10.00-\frac{\sin \left(2\pi \right)}{\left(3.14/5\right)}\right)-\left(9.90-\frac{\sin \left(1.98\pi \right)}{\left(3.14/5\right)}\right)\right]\end{array}$

The value of $\sin \left(1.98\pi \right)$ is $0.108$ and $\sin \left(2\pi \right)$ is $0$.

On further solving the above equation, by substituting the value of $\sin \left(1.01\pi \right)$ and $\sin \left(0.99\pi \right)$.

    $\begin{array}{c}\text{Probability}=\frac{1}{10}\times \left[\left(10.00-\frac{0}{0.628}\right)-\left(9.90-\frac{0.108}{0.628}\right)\right]\\ =\frac{1}{10}\times \left[\left(10.00-0\right)-\left(9.90-0.171\right)\right]\\ =\frac{1}{10}\times \left[\left(10\right)-\left(9.729\right)\right]\\ =\underset{\_}{0.0271}\end{array}$

Hence, the probability that the particle is in between $x=9.90\text{nm}$ and $x=10.00\text{nm}$, is $\underset{\_}{0.0271}$.

(d)

Interpretation Introduction

Interpretation:

The probability of finding the particle of wavefunction ${\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$ between $x=5.00\text{nm}$ and $x=10.00\text{nm}$ has to be calculated.

Concept introduction:

The wave function represents the exact position of the electron in an atom.  For a wave function to be acceptable, it must be normalized.  The wave function is represented by $\psi$ in the quantum mechanics.

Answer to Problem 7B.7P

The probability that the particle is in between $x=5.00\text{nm}$ and $x=10.00\text{nm}$, is $\underset{\_}{0.5}$.

Explanation of Solution

The probability that the particle is in between $x=5.00\text{nm}$ and $x=10.00\text{nm}$, is calculated by the following formula.

    $\text{Probability}={\displaystyle \underset{5.00}{\overset{10.00}{\int }}{\psi }^{\ast }\psi dx}$                                                                               (4)

Where,

• $\psi$ is the normalized wave function.
• ${\psi }^{\ast }$ is the conjugated complex of the normalized wave function.

The given wavefunction of the particle is shown below.

    $\psi ={\left(\frac{2}{L}\right)}^{1/2}\sin \left(\pi x/L\right)$

The wave function is real.  So, ${\psi }^{\ast }$ is equal to $\psi$.

Substitute the value of ${\psi }^{\ast }$ and $\psi$ in equation (4).

    $\begin{array}{c}\text{Probability}={\displaystyle \underset{5.00}{\overset{10.00}{\int }}{\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right){\left(2/L\right)}^{1/2}\sin \left(\pi x/L\right)dx}\\ ={\displaystyle \underset{5.00}{\overset{10.00}{\int }}\left(2/L\right){\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{5.00}{\overset{10.00}{\int }}{\sin }^2\left(\pi x/L\right)dx}\\ =\left(2/L\right){\displaystyle \underset{5.00}{\overset{10.00}{\int }}\frac{1}{2}\left(1-\cos 2\left(\pi x/L\right)\right)dx}\end{array}$

The identity used in the above equation is shown below.

    ${\sin }^{2}x=\frac{1}{2}\left(1-\cos 2x\right)$

On further solving the above integration, by substituting the value of $L=10\text{nm}$ and $\pi =3.14$.

    $\begin{array}{c}\text{Probability}=\frac{1}{2}\left(2/10\right){\displaystyle \underset{5.00}{\overset{10.00}{\int }}\left(1-\cos 2\left(\pi x/10\right)\right)dx}\\ =\frac{1}{10}\times {\left[x-\frac{\sin 2\left(\pi x/10\right)}{2\left(\pi /10\right)}\right]}_{5.00}^{10.00}\\ =\frac{1}{10}\times \left[\left(10.00-\frac{\sin 2\left(10.00\times \pi /10\right)}{2\left(\pi /10\right)}\right)-\left(5.00-\frac{\sin 2\left(5\times \pi /10\right)}{2\left(\pi /10\right)}\right)\right]\\ =\frac{1}{10}\times \left[\left(10.00-\frac{\sin \left(2\pi \right)}{\left(3.14/5\right)}\right)-\left(5.00-\frac{\sin \left(\pi \right)}{\left(3.14/5\right)}\right)\right]\end{array}$

The value of $\sin \left(\pi \right)$ is $0$ and $\sin \left(2\pi \right)$ is $0$.

On further solving the above equation, by substituting the value of $\sin \left(1.01\pi \right)$ and $\sin \left(0.99\pi \right)$.

    $\begin{array}{c}\text{Probability}=\frac{1}{10}\times \left[\left(10.00-\frac{0}{0.628}\right)-\left(5.00-\frac{0}{0.628}\right)\right]\\ =\frac{1}{10}\times \left[10-5\right]\\ =\frac{1}{10}\times \left[5\right]\\ =\underset{\_}{0.5}\end{array}$

Hence, the probability that the particle is in between $x=5.00\text{nm}$ and $x=10.00\text{nm}$, is $\underset{\_}{0.5}$.