Chapter 8, Problem 8P

To determine

The expression for the allowed energies of particle and the ground state energy for an electron.

Answer to Problem 8P

The expression for the allowed energies of particle is ${\left(c^2\frac{{\hslash }^2{\pi }^2}{L^2}\left(n_1^2+n_2^2+n_3^2\right)+m^2{c}^4\right)}^{1/2}$ the ground state energy for an electron is $107.4\text{MeV}$.

Explanation of Solution

Write the expression for momentum given by de Broglie relation.

    $\left|\text{p}\right|=\hslash \left|\text{k}\right|$        (I)

Here, $p$ is the momentum, $\hslash$ is the Planck’s constant and $k$ is the wave vector.

Here, $\text{k}=\left(k_1,k_2,k_3\right)$ and $k_1,k_2\text{and}{k}_3$ are wavenumbers in three mutually perpendicular axes.

Write the expression for energy given by de Broglie relation.

  $E={\left(c^2{\left|p\right|}^2+m^2{c}^4\right)}^{1/2}$        (II)

Here, $E$ is the energy, $c$ is the speed of light and $m$ is the mass.

Assume that particle is a wave and it must vanish at boundaries of the box so the expression for wavelength is:

    ${\lambda }_1=\frac{2\pi }{k_1}$        (III)

Write the expression for length of particle box in terms of wavelength.

    $L=n_1\left(\frac{{\lambda }_1}{2}\right)$        (IV)

From equation (III) and equation (IV), it can be concluded that:    $k_1=\frac{n_1\pi }{L}$

Similarly, $k_2=\frac{n_2\pi }{L}$ and $k_3=\frac{n_3\pi }{L}$ .

Take the square of momentum in equation (I).

    ${\left|\text{p}\right|}^2={\hslash }^2{\left|\text{k}\right|}^2$

Substitute $k_1^2+k_2^2+k_3^2$ for ${\text{k}}^2$ in above equation.

     ${\left|\text{p}\right|}^2={\hslash }^2\left(k_1^2+k_2^2+k_3^2\right)$

Substitute $\frac{n_1\pi }{L}$ for $k_1$, $\frac{n_2\pi }{L}$ for $k_2$ and $\frac{n_3\pi }{L}$ for $k_3$ in above expression.

    $\begin{array}{c}{\left|\text{p}\right|}^2={\hslash }^2\left(\frac{n_1^2{\pi }^2}{L^2}+\frac{n_2^2{\pi }^2}{L^2}+\frac{n_3^2{\pi }^2}{L^2}\right)\\ =\frac{{\hslash }^2{\pi }^2}{L^2}\left(n_1^2+n_2^2+n_3^2\right)\end{array}$

Substitute $\frac{{\hslash }^2{\pi }^2}{L^2}\left(n_1^2+n_2^2+n_3^2\right)$ for ${\left|\text{p}\right|}^2$ in equation (II).

    $E={\left(c^2\frac{{\hslash }^2{\pi }^2}{L^2}\left(n_1^2+n_2^2+n_3^2\right)+m^2{c}^4\right)}^{1/2}$

Conclusion:

When the electron is in the lowest energy state is $n_1=n_2=n_3=1$

Substitute $10\text{fm}$ for $L$ , $1.055\times {10}^{-34}\text{Js}$ for $\hslash$ , $9.1\times {10}^{-31}\text{kg}$ for $m$ and $3\times {10}^8\text{m/s}$ for $c$ in above equation.

    $\begin{array}{c}E={\left(\begin{array}{l}{\left(3\times {10}^8\text{m/s}\right)}^2\frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2{\pi }^2}{\left(10\times {10}^{-15}\text{m}\right)}\left(1+1+1\right)+\\ {\left(9.1\times {10}^{-31}\text{kg}\right)}^2{\left(3\times {10}^8\text{m/s}\right)}^4\end{array}\right)}^{1/2}\\ ={\left(2.965\times {10}^{-22}+6.707\times {10}^{-27}\right)}^{1/2}\text{J}\\ =1.722\times {10}^{-11}\text{J}\left(\frac{6.24\times {10}^{12}\text{MeV}}{1\text{J}}\right)\\ =107.4\text{MeV}\end{array}$

Thus, the expression for the allowed energies of particle is ${\left(c^2\frac{{\hslash }^2{\pi }^2}{L^2}\left(n_1^2+n_2^2+n_3^2\right)+m^2{c}^4\right)}^{1/2}$ the ground state energy for an electron is $107.4\text{MeV}$.