Chapter 5, Problem 14P

(a)

To determine

Show that the formula for low-energy electron diffraction (LEED), when electrons are incident perpendicular to a crystal surface, may be written as $\sin \varphi =\frac{nhc}{d{\left(2m_e{c}^{2}K\right)}^{1/2}}$.

Answer to Problem 14P

It is shown that the formula for low-energy electron diffraction (LEED), when electrons are incident perpendicular to a crystal surface, may be written as $\sin \varphi =\frac{nhc}{d{\left(2m_e{c}^{2}K\right)}^{1/2}}$.

Explanation of Solution

Write the expression for the diffraction condition for a crystal.

    $n\lambda =d\sin \varphi$        (I)

Here, $n$ is the order of diffraction, $\lambda$ is the wavelength, $d$ is the atomic spacing, $\varphi$ is the angle of incidence.

Write the expression for $\lambda$.

    $\lambda =\frac{h}{p}$        (II)

Here, $h$ s the Planck’s constant, $p$ is the linear momentum.

Write the expression for $p$.

    $p={\left(2m_{e}K\right)}^{1/2}$        (III)

Here, $m_e$ is the mass of electron, $K$ is the kinetic energy of the electron.

Use equation (III) in (II) to solve for $\lambda$.

    $\lambda =\frac{h}{{\left(2m_{e}K\right)}^{1/2}}$        (IV)

Use equation (IV) in (I) to solve for $\sin \varphi$.

    $\begin{array}{c}\sin \varphi =\frac{n}{d}\left(\frac{h}{{\left(2m_{e}K\right)}^{1/2}}\right)\\ =\frac{nhc}{d\left({\left(2m_e{c}^{2}K\right)}^{1/2}\right)}\end{array}$        (V)

Conclusion:

Therefore, it is shown that the formula for low-energy electron diffraction (LEED), when electrons are incident perpendicular to a crystal surface, may be written as $\sin \varphi =\frac{nhc}{d{\left(2m_e{c}^{2}K\right)}^{1/2}}$.

(b)

To determine

The atomic spacing in a crystal that has diffraction maxima at $\varphi =24.1°$ and $\varphi =54.9°$ for $100\text{eV}$ electrons.

Answer to Problem 14P

The atomic spacing in a crystal that has diffraction maxima at $\varphi =24.1°$ is $\underset{\_}{3.00\times {10}^{-10}\text{m}}$ and $\varphi =54.9°$ is $\underset{\_}{3.00\times {10}^{-10}\text{m}}$ for $100\text{eV}$ electrons.

Explanation of Solution

Use equation (V) to solve for the atomic spacing for the crystal.

    $d_n=\frac{nhc}{\sin \varphi {\left(2m_{e}K\right)}^{1/2}}$        (VI)

Conclusion:

Substitute $1$ for $n$, $6.636\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $24.1°$ for $\varphi$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$, $100\text{eV}$ for $K$ in equation (VI) to find $d_1$.

    $\begin{array}{c}{d}_1=\frac{\left(1\right)\left(6.636\times {10}^{-34}\text{J}\cdot \text{s}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\sin 24.1°\left[2\left(\left(9.1\times {10}^{-31}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\left(100\text{eV}\right)\right]}\\ =3.00\times {10}^{-10}\text{m}\end{array}$

Substitute $2$ for $n$, $6.636\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m}/\text{s}$ for $c$, $54.9°$ for $\varphi$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$, $100\text{eV}$ for $K$ in equation (VI) to find $d_1$.

    $\begin{array}{c}{d}_1=\frac{\left(2\right)\left(6.636\times {10}^{-34}\text{J}\cdot \text{s}\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\sin 54.9°\left[2\left(\left(9.1\times {10}^{-31}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\times \frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\left(100\text{eV}\right)\right]}\\ =3.00\times {10}^{-10}\text{m}\end{array}$

The atomic spacing in both cases, $24.1°$ corresponds to $n=1$ and $54.9°$ to $n=2$.

Therefore, the atomic spacing in a crystal that has diffraction maxima at $\varphi =24.1°$ is $\underset{\_}{3.00\times {10}^{-10}\text{m}}$ and $\varphi =54.9°$ is $\underset{\_}{3.00\times {10}^{-10}\text{m}}$ for $100\text{eV}$ electrons.