Question

The wavelength of $10keV$ electron beam is $0.1227\overset{^\circ }{A}$ . When these waves are diffracted from a metal foil having $d=0.55\overset{^\circ }{A}$ . 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
$\lambda =\frac{h}{\sqrt{2 mE}}$
$\lambda =\frac{h}{\sqrt{2 meV}}=\sqrt{\frac{150}{V}}\overset{^\circ }{A},\left( for electron M = 9 . 11 \times \left(10\right)^{- 31} kg\right)$
$\lambda =\sqrt{\frac{150}{10 \times 10^{3}}}=0.122\overset{^\circ }{A}$
For minima,
$dsin\theta =n\lambda $
For first minima, $n=1$
$0.55sin\theta =0.122$
$sin\theta =\frac{0 . 122}{0 . 55}=0.2218$