An electron beam of energy 10 , KeV is incide | vedclass

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(A) The de Broglie wavelength $\lambda$ of an electron with energy $V$ (in volts) is given by $\lambda = \sqrt{\frac{150}{V}} \, \mathring{A}$.Given e

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(A) The de Broglie wavelength $\lambda$ of an electron with energy $V$ (in volts) is given by $\lambda = \sqrt{\frac{150}{V}} \, \mathring{A}$.Given energy $E = 10 \, KeV = 10^4 \, eV$,so $V = 10^4 \, V$.$\lambda = \frac{12.27}{\sqrt{10^4}} \, \mathring{A} = \frac{12.27}{100} \, \mathring{A} = 0.1227 \, \mathring{A}$.According to Bragg's law for diffraction,$n \lambda = d \sin \phi$,where $d = 0.2454 \, \mathring{A}$ and assuming $n = 1$ for the first order.$0.1227 = 0.2454 \sin \phi$.$\sin \phi = \frac{0.1227}{0.2454} = 0.5$.$\phi = \arcsin(0.5) = 30^{\circ}$.

An electron beam of energy $10 \, KeV$ is incident on a metallic foil. If the interatomic distance is $0.2454 \, \mathring{A}$,then find the angle of diffraction in degrees. $(\sqrt{150} = 12.27)$

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