Question

The wave properties of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively as shown in fig :

(i). Electrons accelerated by potential V are diffracted from a crystal. If d = 1Aͦ and i = 30°, V should be about (h = 6.6 × 10⁻³⁴Js),

(m_e = 9.1 × 10⁻³¹kg,e = 1.6 × 10⁻¹⁹C) 

(a) 1000V 

(b) 2000V 

(c) 50V 

(d) 500V

(ii). What is the glancing angle in the above experiment? 

(a) 30° 

(b) 15° 

(c) 45° 

(d) 60°

(iii). If a strong diffraction peak is observed when electrons are incident at an angle ‘i’ from the normal to the crystal planes with distance d between them, de- Broglie wavelength λ_dB of electrons can be calculated by the relationship (n is an integer) :

(a) dcosi = nλ_dB 

(b) dsini = nλ_dB 

(c) 2d cos i = nλ_dB 

(d) 2d sin I = nλ_dB

Answer 1

(1) Option : (c) 

Path difference (Δx) between two electron waves reflected from the two planes of the atoms in a crystal is :

∆x = d sin(90 - i) + d sin(90 - i) 

= d cos 30° + d cos 30° 

= 2d cos 30° 

For diffraction of these two waves, 

2d cos 30° = λ

(2) Option : (d) 

Glancing angle is the inclination of the incident rays with the plane of the atoms, 

Which is : 

θ = 90 − 30 = 60°

(3) Option : (c) 

The Bragg’s relation for having an intensity maximum for diffraction pattern is : 

2d sinθ = nλ, 

Where θ is glancing angle 

Hence,

θ = (90° − i) 

∴ 2d sin(90° − i) 

= nλ_dB

Or,