Chapter 40, Problem 67P

(a)

To determine

The wavelength of a particle in the ground state of a one-dimensional infinite square well.

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is $2.00\text{fm}$ .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  $\lambda =2L$   ........ (1)

Here, $\lambda$ is the wavelength of the particle and $L$ is the length of the one-dimensional infinite square well.

Calculation:

Substitute $2.00\text{fm}$ for $L$ in equation (1).

  $\begin{array}{c}\lambda =2\left(2.00\text{fm}\right)\\ =4.00\text{fm}\end{array}$

Conclusion:

Thus, the wavelength of a particle in the ground state of a one-dimensional infinite square well is $4.00\text{fm}$ .

(b)

To determine

The momentum of a particle in the ground state of a one-dimensional infinite square well.

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is $2.00\text{fm}$ .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  $\lambda =2L$   ........ (1)

Here, $\lambda$ is the wavelength of the particle and $L$ is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  $p=\frac{h}{\lambda }$

Here, $p$ is the momentum of the particle and $h$ is the Planck’s constant.

Multiplying numerator and denominator by $c$ in the above expression.

  $p=\frac{hc}{\lambda c}$   ........ (2)

Here, $c$ is the speed of light.

Calculation:

Substitute $2.00\text{fm}$ for $L$ in equation (1).

  $\begin{array}{c}\lambda =2\left(2.00\text{fm}\right)\\ =4.00\text{fm}\end{array}$

Substitute $4.00\text{fm}$ for $\lambda$ and $1240\text{MeV}\cdot \text{fm}$ for $hc$ in equation (2).

  $\begin{array}{l}p=\frac{\left(1240\text{MeV}\cdot \text{fm}\right)}{\left(4.00\text{fm}\right)c}\\ =310\text{MeV}/c\end{array}$

Conclusion:

Thus, the momentum of a particle in the ground state of a one-dimensional infinite square well is $310\text{MeV}/c$ .

(c)

To determine

To show: The total energy of an electron in the ground state of a one-dimensional infinite square well is approximately $pc$ .

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is $2.00\text{fm}$ .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  $\lambda =2L$   ........ (1)

Here, $\lambda$ is the wavelength of the particle and $L$ is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  $p=\frac{h}{\lambda }$

Here, $p$ is the momentum of the particle and $h$ is the Planck’s constant.

Multiplying numerator and denominator by $c$ in the above expression.

  $p=\frac{hc}{\lambda c}$   ........ (2)

Here, $c$ is the speed of light.

Write the expression for the total energy of the electron.

  $\begin{array}{c}{E}^2=E_0{}^2+p^2{c}^2\\ =p^2{c}^2\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)\end{array}$   ........ (3)

Here, $E$ is the total energy of the electron and $E_0$ is the rest energy of the electron.

Calculation:

Substitute $2.00\text{fm}$ for $L$ in equation (1).

  $\begin{array}{c}\lambda =2\left(2.00\text{fm}\right)\\ =4.00\text{fm}\end{array}$

Substitute $4.00\text{fm}$ for $\lambda$ and $1240\text{MeV}\cdot \text{fm}$ for $hc$ in equation (2).

  $\begin{array}{l}p=\frac{\left(1240\text{MeV}\cdot \text{fm}\right)}{\left(4.00\text{fm}\right)c}\\ =310\text{MeV}/c\end{array}$

Substitute $310\text{MeV}/c$ for $p$ in $\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)$ .

  $\begin{array}{c}\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)=\left(1+\frac{E_0{}^2}{{\left(310\text{MeV}/c\right)}^2{c}^2}\right)\\ \Rightarrow E_0{}^2<<p^2{c}^2\text{and}\\ \left(1+\frac{E_0{}^2}{p^2{c}^2}\right)\approx 1\end{array}$

Substitute $1$ for $\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)$ in equation (3).

  $E^2\approx p^2{c}^2$

Simplify the above expression for the total energy of the electron.

  $E\approx pc$

Conclusion:

Thus, the total energy of an electron in the ground state of a one-dimensional infinite square well is approximately $pc$ .

(d)

To determine

The kinetic energy of an electron in the ground state of a one-dimensional infinite square well.

Explanation of Solution

Given:

The length of the one-dimensional infinite square well is $2.00\text{fm}$ .

Formula used:

Write the expression for the wavelength of a particle in the ground state of a one-dimensional infinite square well.

  $\lambda =2L$   ........ (1)

Here, $\lambda$ is the wavelength of the particle and $L$ is the length of the one-dimensional infinite square well.

Write the de Broglie relation for the momentum.

  $p=\frac{h}{\lambda }$

Here, $p$ is the momentum of the particle and $h$ is the Planck’s constant.

Multiplying numerator and denominator by $c$ in the above expression.

  $p=\frac{hc}{\lambda c}$   ........ (2)

Here, $c$ is the speed of light.

Write the expression for the total energy of the electron.

  $\begin{array}{c}{E}^2=E_0{}^2+p^2{c}^2\\ =p^2{c}^2\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)\end{array}$   ........ (3)

Here, $E$ is the total energy of the electron and $E_0$ is the rest energy of the electron.

Calculation:

Substitute $2.00\text{fm}$ for $L$ in equation (1).

  $\begin{array}{c}\lambda =2\left(2.00\text{fm}\right)\\ =4.00\text{fm}\end{array}$

Substitute $4.00\text{fm}$ for $\lambda$ and $1240\text{MeV}\cdot \text{fm}$ for $hc$ in equation (2).

  $\begin{array}{l}p=\frac{\left(1240\text{MeV}\cdot \text{fm}\right)}{\left(4.00\text{fm}\right)c}\\ =310\frac{\text{MeV}}{c}\end{array}$

Substitute $310\text{MeV}/c$ for $p$ in $\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)$ .

  $\begin{array}{c}\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)=\left(1+\frac{E_0{}^2}{{\left(310\frac{\text{MeV}}{c}\right)}^2{c}^2}\right)\\ \Rightarrow E_0{}^2<<p^2{c}^2\text{and}\\ \left(1+\frac{E_0{}^2}{p^2{c}^2}\right)\approx 1\end{array}$

Substitute $1$ for $\left(1+\frac{E_0{}^2}{p^2{c}^2}\right)$ in equation (3).

  $E^2\approx p^2{c}^2$

Simplify the above expression for the total energy of the electron.

  $E\approx pc$

Write the expression for the kinetic energy of an electron in the ground state of a one-dimensional infinite square well.

  $\begin{array}{c}K=E-E_0\\ \approx E\end{array}$

Here, $K$ is the kinetic energy of an electron in the ground state of the well.

Substitute $pc$ for $E$ in the above expression.

  $K=pc$   ........ (4)

Substitute $310\text{MeV}/c$ for $p$ in equation (4).

  $\begin{array}{c}K=\left(310\frac{\text{MeV}}{c}\right)c\\ =310\text{MeV}\end{array}$

Conclusion:

Thus, the kinetic energy of an electron in the ground state of a one-dimensional infinite square well is $310\text{MeV}$ .