Question

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to $1.54 Å$. Mass of electron $=9 \cdot 1 \times 10^{-31} kg , h=6 \cdot 62 \times 10^{-34} Js$.

Answer 1

[Method : By de Broglie relationship we calculate velocity of electron, and its kinctic cnergy in clectron-volt, and hence the value of potential]

$\begin{array}{rlr}\lambda & =\frac{h}{m V} ; \quad \text { and } \quad 1.54 Å=1.54 \times 10^{-10} m \\v & =\frac{h}{m \lambda}=\frac{6.62 \times 10^{-34}}{9 \cdot 1 \times 10^{-31} \times 1.54 \times 10^{-10}} ms ^{-1}=0 \cdot 47 \times 10^{7} ms ^{-1} \\KE & =\frac{1}{2} m v^{2}=\frac{1}{2} \times 9 \cdot 1 \times 10^{-31} \times(0 \cdot 47)^{2} \times 10^{14} J \\& =1 \cdot 0 \times 10^{-17} J \quad\left(1 eV =1.60 \times 10^{-19} J \right) \\& =\frac{1.0 \times 10^{-17}}{1.60 \times 10^{-19}} cV =0.625 \times 10^{2} eV =62.5 cV\end{array}$

Potential through which the electron should be accelerated $=62 \cdot 5 V$.