Question

Wave property of electrons implies that they will show diffraction effects. Davisson-Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively.
Electrons accelerated by potential $V$ are diffracted from a crystal. If $d=1\mathring{\text{A}}$ and $i={30}^{\circ }$ then $V$ should be about $(h=6.6\times {10}^{-34}\text{Js},m=9.1\times {10}^{31}\text{kg},e=1.6\times {10}^{-19}\text{C})$

• A. $500\text{V}$
• B. $1000\text{V}$
  
• C. $2000\text{V}$
  
• D. $50\text{V}$

Answer 1

The correct option is D $50\text{V}$

Given, $d=1\mathring{A};i={30}^{\circ }h=6.6\times {10}^{-34}\text{Js};m=9.1\times {10}^{31}\text{kg}e=1.6\times {10}^{-19}\text{C}$



For constructive interference, path difference must be,

$\Delta x=n\lambda$

From the figure, we

Path difference, $\Delta x=QP+PR$

$=OP\cos i+OP\cos i=2d\cos i[\because OP=d]$

$\therefore 2d\cos i=n\lambda$

$\Rightarrow \lambda =\frac{2d\cos i}{n}=\frac{2\times 1\times {10}^{-10}\times \cos {30}^{\circ }}{n}=\frac{\sqrt{3}}{n}\mathring{A}$

For $n=1\Rightarrow \lambda =\sqrt{3}\mathring{A}$

The de-Broglie wavelength associated with an electron acceleration through a potential difference $V$ is,

$\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}$

$\Rightarrow V=\frac{(12.27{)}^2}{(\lambda {)}^2}=\frac{150.55}{3}\approx 50\text{V}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.

Why this question? Key formula: The de-Broglie wavelength associated with an electron acceleration through a potential difference $V$ is, $\lambda =\frac{12.27}{\sqrt{V}}\mathring{A}$