Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy, an X-ray photon or the electron? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of interatomic spacing in the lattice) (m_e = 9.11 x 10^–31 kg).

Solution

Wavelength of the probe = 1 $\overset{o}{A}$
Mass of electron, m_e = 9.11 x 10^–31 kg

Energy of photon, $E = hv = \frac{hc}{λ}$
   $= \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{10^{- 10}} J = 19.89 \times 10^{- 16} J$
i.e., $E = \frac{19.89 \times 10^{- 16}}{1.6 \times 10^{- 19}} eV = 12.43 keV$

For the case of electron,

   $λ = \frac{h}{\sqrt{2 mE}}$

$\Rightarrow$ $\sqrt{2 mE} = \frac{h}{λ}$

$\Rightarrow$ $2 mE = \frac{h^{2}}{λ^{2}}$
$\Rightarrow$    $E = \frac{h^{2}}{2 {mλ}^{2}}$
$\Rightarrow$ $E = \frac{{(6.63 \times 10^{- 34})}^{2}}{2 \times 9.11 \times 10^{- 31} \times 10^{- 20}} J$
$= \frac{(6.63 \times 10^{- 34})^{2}}{2 \times 9.11 \times 10^{- 31} \times 10^{- 20} \times 1.6 \times 10^{- 19}} eV$
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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