Question

Question: Atoms in a crystal form atomic planes at many different angles with respect to the surface. The accompanying figure shows the behaviors of representative incident and scattered waves in the Davisson-Germer experiment. A beam of electrons accelerated through 54 V is directed normally at a nickel surface, and strong reflection is detected only at an angle $\varphi$of 50⁰.Using the Bragg law, show that this implies a spacing D of nickel atoms on the surface in agreement with the known value of 0.22 nm.

Short answer

Answer:

It is shown that D = 0.22nm .

Step-by-step solution

de Broglie’s formula

The de Broglie’s wavelength formula is $\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}}{\mathbf{m}\mathbf{v}}$.

Proof

Using the formula , $V=\frac{h^2}{2mq{\lambda }^2}$find the value of as follows:

$$\begin{gathered} 54V=\frac{{\left(6.63\times {10}^{-34}J·s\right)}^2}{2\left(9.11\times {10}^{-31}kg\right)\left(1.6\times {10}^{-19}C\right){\lambda }^2} \\ {\lambda }^2=\frac{{\left(6.63\times {10}^{-34}J·s\right)}^2}{108V\left(9.11\times {10}^{-31}kg\right)\left(1.6\times {10}^{-19}C\right)} \\ \lambda =\frac{6.63\times {10}^{-34}J·s}{\sqrt{108V\left(9.11\times {10}^{-31}kg\right)\left(1.6\times {10}^{-19}C\right)}} \\ \lambda =1.67\times {10}^{-10} \text{m} \end{gathered}$$

From the figure, it is clear that the angle between the incident beam and atomic plate can be obtained as follows:

$$\begin{gathered} \theta =90°-\frac{\varphi }{2} \\ =90°-\frac{50°}{2} \\ =90°-25° \\ =65° \end{gathered}$$

Using Bragg’s law, we have:

$$\begin{gathered} 2d\sin \theta =m\lambda \\ d=\frac{m\lambda }{2\sin \theta } \\ d=\frac{1\times 1.67\times {10}^{-10}m}{2\times \sin 65°} \\ d=0.092 \text{nm} \end{gathered}$$

The angle between the planes and the crystal surface can be obtained as follows:

$$\begin{gathered} \theta =90°-65° \\ =25° \end{gathered}$$

Now, by the sine law, we get:

$$\begin{gathered} \sin \theta =\frac{d}{D} \\ D=\frac{d}{\sin \theta } \\ D=\frac{0.092nm}{\sin 25°} \\ D=0.22 \text{nm} \end{gathered}$$

Thus, it is proved that spacing has the value D of 0.22nm.