Question

The wavelength of $10keV$ electron beam is $0.1227\stackrel{{}^{\circ }}{A}$ . When these waves are diffracted from a metal foil having $d=0.55\stackrel{{}^{\circ }}{A}$ . At angle $\theta$ first maxima will be present, then what will be the value of $sin\theta$ :

• A. $0.1116$
• B. $0.2218$
• C. $0.4464$
• D. $0.8928$

Answer 1

Answer: (B)

The wavelength associated with a particle is given by
$\lambda =\frac{h}{\sqrt{2mE}}$
$\lambda =\frac{h}{\sqrt{2meV}}=\sqrt{\frac{150}{V}}\stackrel{{}^{\circ }}{A},\left(forelectronM=9.11\times {\left(10\right)}^{-31}kg\right)$
$\lambda =\sqrt{\frac{150}{10\times 10^3}}=0.122\stackrel{{}^{\circ }}{A}$
For minima,
$dsin\theta =n\lambda$
For first minima, $n=1$
$0.55sin\theta =0.122$
$sin\theta =\frac{0.122}{0.55}=0.2218$