Question

Electrons accelerated by potential V are diffracted from a crystal. If d= $1\mathring{A}$ and $i-{30}^o$, V should be about $(h=6.6\times {10}^{-34}Js,m_e=9.1\times {10}^{-31}kg,e=1.6\times {10}^{-19}C)$

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

Answer 1

The correct option is B 50 V

For constructive interference, $2dcosi=n\lambda$ and also wavelength, $\lambda =\frac{h}{\sqrt{2meV}}$

So, $2dcosi=n\times \frac{h}{\sqrt{2meV}}$

Squaring both sides, $4d^{2}co{s}^{2}i=\frac{n^2{h}^2}{2meV}$

$V=\frac{n^2{h}^2}{8d^{2}meco{s}^{2}i}=\frac{\left(1{)}^2\right(6.6\times {10}^{-34}{)}^2}{8(1\times {10}^{-10}{)}^2(9.1\times {10}^{-31}\left)\right(1.6\times {10}^{-19}\left)\right(co{s}^{2}30)}=49.8V\sim 50V$ (we assume first order interference and take $n=1$)