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) (me=9.11 × 10⁻³¹ kg).

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

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

Mass of an electron, m_e = 9.11 x 10^-31 kg

Planck’s constant, h = 6.6 x 10⁻³⁴ Js

Charge on an electron, e = 1.6 x 10^-19 C

The kinetic energy of the electron is given as:

E = 1/2 m_ev²

=>m_ev  = √ (2Em_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:

λ = h/p = h/m_ev = h/√ (2Em_e)

Therefore , E = h²/(2λ²m_e) 2

= (6.6 x 10⁻³⁴)² /2 x ( 10⁻¹⁰)² x(9.11 x 10^-31 ) = 2.39 x 10⁻¹⁷ J

= ( 2.39 x 10⁻¹⁷ )/(1.6 x 10^-19) = 149 .375 eV

Energy of a photon,E’ =hc/λ e  eV

= (6.6 x 10⁻³⁴) x (3 x 10⁸ ) /( 10⁻¹⁰) (1.6 x 10^-19)

= 12.375 x  10³  eV =12.375 keV

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