Electrons are accelerated through 450 V and incident on a graphite crystal whose lattice planes have spacing 0.210 nm. Using Bragg’s law for electron diffraction, determine the angle at which the second‑order diffraction maximum is observed.

hard 3 marks Class 12 · Physics · Dual Nature of Radiation and Matter

Answer

16.0°

The correct answer is 16.0° — Insert $d = 0.210$  nm: $s in θ = 0.0578/0.210 = 0.2754$ and $θ = s i n^{- 1} (0.2754) = 16. 0^{\circ}$ .

Step-by-step solution

Find kinetic energy: $K = e V = 1.602 \times 1 0^{- 19} \times 450 = 7.209 \times 1 0^{- 17}$  J.
Compute momentum: $p = 2 m K = 1.146 \times 1 0^{- 23}$  kg m s⁻¹.
Find de Broglie wavelength: $λ = h / p = 6.626 \times 1 0^{- 34} /1.146 \times 1 0^{- 23} = 5.78 \times 1 0^{- 11}$  m = 0.0578 nm.
Apply Bragg’s law for second order ( $n = 2$ ): $2 λ = 2 d s in θ$ , hence $s in θ = λ / d$ .
Insert $d = 0.210$  nm: $s in θ = 0.0578/0.210 = 0.2754$ and $θ = s i n^{- 1} (0.2754) = 16. 0^{\circ}$ .

Common mistake: Using first‐order equation instead of

n = 2

, or mixing units (nm vs m) in the calculation

Concept tested: Bragg diffraction of electrons