Chapter 6, Problem 17P

(a)

To determine

The sketch of wave functions and probability density for $n=1$ and $n=2$ states.

Answer to Problem 17P

The sketch of wave functions and probability density for $n=1$ and $n=2$ states is plotted below.

Explanation of Solution

The sketch of wave function is plotted below.

The sketch of probability density is plotted below.

Conclusion:

The sketch of wave functions and probability density for $n=1$ and $n=2$ states is plotted above.

(b)

To determine

The probability of finding the electron between 0.15 nm and 0.35 nm for $n=1$ state.

Answer to Problem 17P

The probability of finding the electron for $n=1$ state is $0.200$ .

Explanation of Solution

Write the expression for wave function.

  $\psi \left(x\right)=\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$        (I)

Here, $A$ is normalization constant, $x$ is position of box, $n$ is energy state, $L$ is the length of box and $\psi \left(x\right)$ is the wave function.

Write the expression for probability.

  $P={\displaystyle \underset{0}{\overset{L/3}{\int }}{\left|\psi \right|}^{2}dx}$        (II)

Conclusion:

Substitute $\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$ for $\psi \left(x\right)$  and substitute $1$ for $n$ in equation (II).

  $P=\left(\frac{2}{L}\right){\displaystyle \underset{L_i}{\overset{L_f}{\int }}{\sin }^2\left(\frac{\pi x}{L}\right)}dx$

Substitute $1.5\stackrel{\text{o}}{\text{A}}$ for $L_i$ , $3.5\stackrel{\text{o}}{\text{A}}$ for $L_f$ and $10\stackrel{\text{o}}{\text{A}}$ for $L$ in above equation and simplify.

  $\begin{array}{c}P=\frac{1}{10}{\left[x-\frac{5}{\pi }\sin \left(\frac{\pi x}{5}\right)\right]}_{1.5}^{3.5}\\ =\frac{1}{5}\left\{3.5-\frac{5}{\pi }\sin \left[\frac{\pi \left(3.5\right)}{5}\right]-1.5+\frac{5}{\pi }\sin \left[\frac{\pi \left(1.5\right)}{5}\right]\right\}\\ =0.200\end{array}$

Thus, the probability of finding the electron for $n=1$ state is $0.200$ .

(c)

To determine

The probability of finding the electron between 0.15 nm and 0.35 nm for $n=1$ state.

Answer to Problem 17P

The probability of finding the electron for $n=2$ state is $0.351$ .

Explanation of Solution

Conclusion:

Substitute $\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)$ for $\psi \left(x\right)$  and substitute $2$ for $n$ in equation (II).

  $P=\left(\frac{2}{L}\right){\displaystyle \underset{L_i}{\overset{L_f}{\int }}{\sin }^2\left(\frac{2\pi x}{L}\right)}dx$

Substitute $1.5\stackrel{\text{o}}{\text{A}}$ for $L_i$ , $3.5\stackrel{\text{o}}{\text{A}}$ for $L_f$ and $10\stackrel{\text{o}}{\text{A}}$ for $L$ in above equation and simplify.

  $\begin{array}{c}P=\frac{1}{10}{\left[x-\frac{5}{2\pi }\sin \left(0.4\pi x\right)\right]}_{1.5}^{3.5}\\ =\frac{1}{5}\left\{3.5-\frac{5}{2\pi }\sin \left(1.4\pi \right)-1.5+\frac{5}{2\pi }\sin \left[0.6\pi \right]\right\}\\ =0.351\end{array}$

Thus, the probability of finding the electron for $n=2$ state is $0.351$ .

(d)

To determine

The energies in electron volts for $n=1$ and $n=2$ states.

Answer to Problem 17P

The energy for $n=1$ state is $37.7\text{eV}$ and the energy for $n=2$ state is $151\text{eV}$ .

Explanation of Solution

Write the expression for energy of particle.

  $E=\frac{n^2{h}^2}{8m{L}^2}$        (III)

Here, $n$ is the energy level, $h$ is the Planck’s constant, $m$ is the mass of particle, $L$ is the length of box and $E$ is the energy of particle.

Conclusion:

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ , ${10}^{-10}\text{m}$ for $L$, $9.11\times {10}^{-31}\text{kg}$ for $m$ and $1$ for $n$ in equation (III).

  $\begin{array}{c}{E}_1=\frac{{\left(1\right)}^2{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(9.11\times {10}^{-31}\text{kg}\right){\left({10}^{-10}\text{m}\right)}^2}\\ =37.7\text{eV}\end{array}$

Substitute $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ , ${10}^{-10}\text{m}$ for $L$, $9.11\times {10}^{-31}\text{kg}$ for $m$ and $2$ for $n$ in equation (III).

  $\begin{array}{c}{E}_2=\frac{{\left(2\right)}^2{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(9.11\times {10}^{-31}\text{kg}\right){\left({10}^{-10}\text{m}\right)}^2}\\ =151\text{eV}\end{array}$

Thus, the energy for $n=1$ state is $37.7\text{eV}$ and the energy for $n=2$ state is $151\text{eV}$ .