Electrons accelerated through a potential difference $V$ are diffracted from a crystal whose planes are $d = 1$ Å apart. A strong diffraction peak is observed when the electrons are incident at an angle $i = 30°$ from the normal to the crystal planes, with the condition $2d\cos i = n\lambda$ ($n = 1$). $V$ should be about ($h = 6.6 \times 10^{-34}$ J s, $m = 9.1 \times 10^{-31}$ kg, $e = 1.6 \times 10^{-19}$ C)
Answer: (B) $50$ V
Concept. The Bragg-type condition gives $\lambda$; then $V=\dfrac{h^2}{2me\lambda^2}$.
Step 1. Wavelength. $\lambda=2\times1\times\cos30^\circ=\sqrt3$ Å $=1.73\times10^{-10}$ m.
Step 2. Voltage. $V=\dfrac{(6.6\times10^{-34})^2}{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times3\times10^{-20}}=50$ V.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
