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  • Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy?

Q: 11.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy?  (For quantitative comparison, take the wavelength of the probe equal to 1\hspace{1mm}\dot{A}, which is of the order of inter-atomic spacing in the lattice) (m_e=9.11\times 10^-^3^1\hspace{1mm}kg).

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According to De Broglie's equation

p=\frac{h}{\lambda }

The kinetic energy of an electron with De Broglie wavelength 1\hspace{1mm}\dot{A} is given by

\\K=\frac{p^{2}}{2m_{e}}\\ K=\frac{h^{2}}{\lambda ^{2}2m_{e}} \\K=\frac{(6.62\times 10^{-34})^{2}}{2\times 10^{-20}\times 9.11\times 10^{-31}\times 1.6\times 10^{-19}}\\ K=149.375\ eV

The kinetic energy of photon having wavelength 1\hspace{1mm}\dot{A} is 

\\E=\frac{hc}{\lambda }\\ E=\frac{6.62\times 10^{-34}\times 3\times 10^{8}}{10^{-10}\times 1.6\times 10^{-19}}\\ E=12.375keV

Therefore for the given wavelength, a photon has much higher energy than an electron.

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