Chapter 41, Problem 61CP

(a)

To determine

The probability of finding an electron in $n=1$ state.

Answer to Problem 61CP

The probability of finding an electron in $n=1$ state is $0.200$.

Explanation of Solution

A quantum dot is modeled as a one-dimensional box with the length of $1.00\text{ nm}$. Here, the given limits to find the probability of finding the electron is between $x_1=0.150\text{ m}$ and $x_2=0.350\text{ nm}$ for $n=1$ state.

Write to formula to find the probability of finding the electron

    $P_n={\displaystyle {\int }_{x_1}^{x_2}{\left|{\psi }_n\right|}^{2}dx}$                                                                         (I)

Here, $P_n$ is the probability of finding the electron in $n^{th}$ state, ${\Psi }_n$ is the wavefunction of $n$ state and $x_1\text{ and }{x}_2$ are the limits.

Write the formula to find the wave function

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

Here, $L$ is the length of the box.

Conclusion:

Substitute equation (II) in (I) and solve

$P_n={\displaystyle {\int }_{x_1}^{x_2}{\left[\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right)\right]}^{2}dx}$

$P_n=\frac{2}{L}{\displaystyle {\int }_{x_1}^{x_2}{\sin }^2\left(\frac{n\pi x}{L}\right)dx}$                                                                  (III)

Substitute $1$ for $n$, $0.150\text{ nm}$ for $x_1$, $0.350\text{ nm}$ for $x_2$, $1.00\text{ nm}$ for $L$ in equation (III) to find the value of $P_1$

$P_1=\left(\frac{2}{1.00\text{ nm}}\right){\displaystyle \underset{0.150}{\overset{0.350}{\int }}{\text{sin}}^2\left(\frac{\pi x}{1.00\text{ nm}}\right)dx}$

Note: $\left[{\displaystyle \int {\sin }^2\left(ax\right)dx}=\left(\frac{x}{2}\right)-\left(\frac{1}{4a}\right)\sin \left(2ax\right)\right]$, using this in the above equation and solving

$\begin{array}{c}{P}_1=\left(2.00/\text{nm}\right){\left[\frac{x}{2}-\frac{1.00\text{ nm}}{\text{4}\pi }\text{sin}\left(\frac{2\pi x}{1.00\text{ nm}}\right)\right]}_{0.150\text{ nm}}^{\text{0}\text{.350 nm}}\\ =\left(1.00/\text{nm}\right){\left[x-\frac{1.00\text{ nm}}{\text{2}\pi }\sin \left(\frac{2\pi x}{1.00\text{ nm}}\right)\right]}_{0.150\text{ nm}}^{\text{0}\text{.350 nm}}\\ =\left(1.00/\text{nm}\right)\{\left(0.350\text{ nm}-0.150\text{ nm}\right)-\frac{1.00\text{ nm}}{\text{2}\pi }\left[\sin \left(0.700\pi \right)-\sin \left(0.300\pi \right)\right]\}\\ =0.200\end{array}$

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

(b)

To determine

The probability of finding an electron in $n=2$ state.

Answer to Problem 61CP

The probability of finding an electron in $n=2$ state is $0.351$.

Explanation of Solution

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

Substitute $2$ for $n$, $0.150\text{ nm}$ for $x_1$, $0.350\text{ nm}$ for $x_2$, $1.00\text{ nm}$ for $L$ in equation (III) to find the value of $P_2$

$P_2=\left(\frac{2}{1.00\text{ nm}}\right){\displaystyle \underset{0.150}{\overset{0.350}{\int }}{\text{sin}}^2\left(\frac{2\pi x}{1.00\text{ nm}}\right)dx}$

Note: $\left[{\displaystyle \int {\sin }^2\left(ax\right)dx}=\left(\frac{x}{2}\right)-\left(\frac{1}{4a}\right)\sin \left(2ax\right)\right]$, using this in the above equation and solving

$\begin{array}{c}{P}_2=\left(2.00/\text{nm}\right){\left[\frac{x}{2}-\frac{1.00\text{ nm}}{8\pi }\text{sin}\left(\frac{4\pi x}{1.00\text{ nm}}\right)\right]}_{0.150\text{ nm}}^{\text{0}\text{.350 nm}}\\ =\left(1.00/\text{nm}\right){\left[x-\frac{1.00\text{ nm}}{4\pi }\sin \left(\frac{4\pi x}{1.00\text{ nm}}\right)\right]}_{0.150\text{ nm}}^{\text{0}\text{.350 nm}}\\ =\left(1.00/\text{nm}\right)\{\left(0.350\text{ nm}-0.150\text{ nm}\right)-\frac{1.00\text{ nm}}{4\pi }\left[\sin \left(1.40\pi \right)-\sin \left(0.600\pi \right)\right]\}\\ =0.351\end{array}$

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

(c)

To determine

The energy of $n=1$ state in electron volt.

Answer to Problem 61CP

The energy of $n=1$ state in electron volt is $0.377\text{ eV}$.

Explanation of Solution

The mass of the electron is $9.11\times {10}^{-31}\text{ kg}$ and length of the quantum box is $1.00\text{ nm}$.

Write the formula to find the energy of $n^{th}$ state

    $E_n=\frac{n^2{h}^2}{8m{L}^2}$                                                                 (IV)

Here, $h$ is Planck’s constant $\left[h=6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right]$, $m$ is the mass of the particle and $E_n$ is the $n^{th}$ state energy.

Conclusion:

Substitute $1$ for $n$, $1.00\text{ nm}$ for $L$, $9.11\times {10}^{-31}\text{ kg}$ for $m$ and  $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ in equation (IV) to find value of $E_1$

$\begin{array}{c}{E}_1=\frac{1^2\times {\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\text{ kg}\right){\left(1.00\text{ nm}\right)}^2}\\ =0.377\text{ eV}\end{array}$

Thus, the energy of $n=1$ state in electron volt is $0.377\text{ eV}$.

(d)

To determine

The energy of $n=2$ state in electron volt.

Answer to Problem 61CP

The energy of $n=2$ state in electron volt is $1.57\text{ eV}$.

Explanation of Solution

Substitute $2$ for $n$, $1.00\text{ nm}$ for $L$, $9.11\times {10}^{-31}\text{ kg}$ for $m$ and  $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ in equation (IV) to find value of $E_2$

$\begin{array}{c}{E}_2=\frac{2^2\times {\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)}^2}{8\times \left(9.11\times {10}^{-31}\text{ kg}\right){\left(1.00\text{ nm}\right)}^2}\\ =1.57\text{ eV}\end{array}$

Thus, the energy of $n=2$ state in electron volt is $1.57\text{ eV}$.