Crystal diffraction experiments can be performed using X – rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (Take wavelength of probe = $1 A^{\circ}$ )

$\begin {array} {1 1} (a)\;X-ray\: probe & \quad (b)\;Electron\: probe \\ (c)\;Both\: have\: equal\: energies & \quad (d)\;Can’t \: be\: determined \end {array}$

Ans : (a)
$\lambda = 1A^{\circ}$
$E=\large\frac{1}{2} m_ev^2$
$\Rightarrow \: m_ev = \sqrt 2Em_e\: \: \: \: \: where,\: v=$ velocity of electron; p(momentum) = $m_e$
So,$\lambda = \large\frac{h}{p} = \large\frac{h}{m_ev} = \large\frac{h}{\sqrt2 Em_e}$
So, $E = \large\frac{h^2}{2 \lambda^2m_e}$
$= \large\frac{(6.6 \times 10^{-34} )^2}{ [2 \times (10^{-10} )^2 \times 9.11 \times 10^{-31} ]}$
$= \large\frac{2.39 \times 10^{-17} J}{1.6 \times 10^{-19} } = 149.375 \: eV$
$E’$ = energy of a photon = $\large\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{ (10^{-10} \times 1.6 \times 10^{-19 } )}$
$= 12.375 \times 10^3\: eV = 12.375\: keV$
So, $E’>E$, i.e, X-ray probe has greater energy than electron probe.

edited Mar 13, 2014