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) (me =9.11 × 10–31 kg).

Answer 1

Given: The wavelength of light emitted from the probe is 1  A 0 .

The kinetic energy of an electron is given as,

E= 1 2 m e v 2 (1)

Where, m e is the mass of the electron and v is the velocity of the electron.

de Broglie wavelength is given as,

λ= h p = h m e v v= h m e λ

Where, h is the Planck’s constant and p is the momentum.

By substituting the value of v in equation (1), we get

E= h 2 2 λ 2 m e

By substituting the given values in the above equation, we get

E= ( 6.6× 10 −34 ) 2 2× ( 10 −10 ) 2 ×9.1× 10 −31 =2.39× 10 −17  J× 1 eV 1.6× 10 −19  J =149.375 eV

Energy of a photon is given as,

E ′ = hc λ

By substituting the given values in the above equation, we get

E ′ = 6.6× 10 −34 ×3× 10 8 10 −10  J× 1 eV 1.6× 10 −19  J =1.2375× 10 4  eV

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