Crystal diffraction experiments can be performed using X-rays, or… - Vidyadip

Question

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 Å, which is of the order of inter-atomic spacing in the lattice) $(m^e = 9.11 \times 10^{–31} kg).$

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Answer

Wavelength of the probe $=1\mathring{\text{A}}$
Mass of electron. me $= 9.11 \times 10^{-31} kg$
Energy of photon, $\text{E}=\text{hv}=\frac{\text{hc}}{\lambda}$
$=\frac{{6.63}\times10^{-34}\times3\times10^8}{10^{-10}}\text{J}$
$= 19.89 \times 10^{-16} J$
$\text{i.e., E}=\frac{19.89\times10^{-16}}{1.6\times10^{-19}}\text{eV}=12.43\ \text{keV}$
For the case of electron,
$\lambda=\frac{\text{h}}{\sqrt{2\text{mE}}}$
$\Rightarrow\ \ \sqrt{2\text{mE}}=\frac{\text{h}}{\lambda}$
$\Rightarrow\ \ 2\text{mE}=\frac{\text{h}^2}{\lambda^2}$
$\Rightarrow\ \ \text{E}=\frac{\text{h}^2}{\text{2m}\lambda^2}$
$\Rightarrow\ \ \text{E}=\frac{(6.63\times10^{-34})^2}{2\times9.11\times10^{-31}\times10^{-20}}\text{J}$
$=\frac{(6.63\times10^{-34})^2}{2\times9.11\times10^{-31}\times10^{-20}\times1.6\times10^{-19}}\text{eV}$
i.e., E = 150.8 eV
From the above calculations, we can see that, for the same given wavelength, kinetic energy of a photon is much greater than that of electron.

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