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).
Given:
Wavelength of probe, λ = 1 Å = 10-10m
Mass of electron, me = 9.11 × 10-31Kg
The kinetic energy of an electron is given by:
We can write,
mv = (2 × m × KE)0.5
P = (2 × m × KE)0.5 ...(1)
Where,
m = mass of electron
v = velocity of electron
p = momentum of particle
De-broglie wavelength is given by,
Where,
λ = Wavelength
h = Planck’s constant
p = momentum
From equation (1) we can write,
…(2)
From equation (2) we can write that,
KE = 
Putting the values in above equation we get,
KE = 
KE = 2.39 × 10-17 J
KE for an electron is 2.39 × 10-17 J.
Now for photon,
E’ = 
…(4)
Where,
E = energy of photon
h = Planck’s constant
c = speed of light
λ = wavelength of the photon
By putting values in equation (4 ) we get,
E’ = 19.6 × 10-16 J
The photon will have more energy as compared to the accelerated electron for the same wavelength.