Question

Wave property of electrons implies that they will show diffraction effects. Davisson and 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 (see figure).
Electrons accelerated by potential V are diffracted from a crystal. If $d=1A^o$ and $i={30}^{\circ }$ , V should be about :
$(h=6.6\times {10}^{-34}js,m=9.1\times {10}^{-31}kg,e=1.6\times {10}^{-19}c)$

• A. $2000$ V
• B. $50$ V
• C. $500$ V
• D. $1000$ V

Answer 1

The correct option is B $50$ V

We have,
$2dsin\theta =n\lambda$
angle of incidence = $90-i$
Given $i={30}^o$, $d=1\times {10}^{-10}m$, $h=6.6\times {10}^{-34}$, $m=9.1\times {10}^{-31}$ and $e=1.6\times {10}^{-19}$
Also $n=1$ for $1^{st}$ order diffraction.
Therefore,
$2\times 1\times {10}^{-10}sin(90-30)=1\times \lambda$
$⟹\lambda =2\times {10}^{-10}\frac{\sqrt{3}}{2}=\sqrt{3}\times {10}^{-10}m$
We have de-Broglie wavelength, $\lambda =\frac{h}{mv}$
and also $\frac{1}{2}m{v}^2=eV$ $⟹v=\sqrt{\frac{2eV}{m}}$
Substituting in de-Broglie wavength, we get,
$\lambda =\frac{h}{\sqrt{2meV}}$= $\sqrt{3}\times {10}^{-10}$m
$⟹V=49.86V$