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 × 10⁻³¹ kg).

Answer 1

An X-ray probe has greater energy than an electron probe for the same wavelength.

Wavelength of light emitted from the probe, λ = 1 Å = 10⁻¹⁰ m

Mass of an electron, m_e = 9.11 × 10⁻³¹ kg

Planck’s constant, h = 6.6 × 10⁻³⁴ Js

Charge on an electron, e = 1.6 × 10⁻¹⁹ C

The kinetic energy of the electron is given as:

${\displaystyle \text{E}=\frac{1}{2}{\text{m}}_{\text{e}}{\text{v}}^2}$

${\displaystyle {\text{m}}_{\text{e}}\text{v}=\sqrt{2{\text{Em}}_{\text{e}}}}$

Where,

v = Velocity of the electron

m_ev = Momentum (p) of the electron

According to the de Broglie principle, the de Broglie wavelength is given as:

${\displaystyle \lambda =\frac{\text{h}}{\text{p}}=\frac{\text{h}}{{\text{m}}_{\text{e}}\text{v}}=\frac{\text{h}}{\sqrt{2{\text{Em}}_{\text{e}}}}}$

∴ E = ${\displaystyle \frac{{\text{h}}^2}{2{\lambda }^2{\text{m}}_{\text{e}}}}$

${\displaystyle =\frac{{(6.6\times {10}^{-34})}^2}{2\times {\left({10}^{-10}\right)}^2\times 9.11\times {10}^{-31}}}$

= 2.39 × 10⁻¹⁷ J

= ${\displaystyle \frac{2.39\times {10}^{-17}}{1.6\times {10}^{-19}}}$

= 149.375 eV

Energy of a photon ${\displaystyle \text{E'}=\frac{\text{hc}}{\lambda \text{e}}\text{eV}}$

= ${\displaystyle \frac{6.6\times {10}^{-34}\times 3\times {10}^8}{{10}^{-10}\times 1.6\times {10}^{-19}}}$

= 12.375 × 10³ eV

= 12.375 keV

Hence, a photon has greater energy than an electron for the same wavelength.