X-ray diffraction — Bragg's law 2d sin θ = nλ and applications

Question

X-rays of wavelength 1.54 A are diffracted by a crystal. The first-order diffraction maximum is observed at an angle of 25.2°. Find the interplanar spacing of the crystal using Bragg’s law. What angle would give the second-order maximum?

(JEE Main 2023, similar pattern)

Solution — Step by Step

When X-rays hit a crystal, constructive interference occurs when:

2d\sin\theta = n\lambda

where $d$ is the interplanar spacing, $\theta$ is the glancing angle (angle with the crystal plane, not the normal), $\lambda$ is the wavelength, and $n$ is the order of diffraction.

For $n = 1$ , $\lambda = 1.54$ A $= 1.54 \times 10^{-10}$ m, $\theta_1 = 25.2°$ :

d = \frac{n\lambda}{2\sin\theta} = \frac{1 \times 1.54 \times 10^{-10}}{2 \times \sin 25.2°}

d = \frac{1.54 \times 10^{-10}}{2 \times 0.426} = \frac{1.54 \times 10^{-10}}{0.852}

\boxed{d \approx 1.81 \times 10^{-10} \text{ m} = 1.81 \text{ A}}

For $n = 2$ :

\sin\theta_2 = \frac{2\lambda}{2d} = \frac{\lambda}{d} = \frac{1.54}{1.81} = 0.851

\boxed{\theta_2 = \sin^{-1}(0.851) \approx 58.3°}

For $n = 3$ : $\sin\theta_3 = 3\lambda/(2d) = 3 \times 0.426 = 1.278 > 1$ . This is impossible, so no third-order maximum exists for this crystal and wavelength combination. The maximum order is $n_{max} = \lfloor 2d/\lambda \rfloor = 2$ .

Why This Works

Crystal planes act like partially reflecting mirrors for X-rays. Rays reflected from successive planes travel different path lengths. The path difference is $2d\sin\theta$ — the factor of 2 comes from the ray going down to the next plane and coming back up. When this path difference equals a whole number of wavelengths, the reflected rays interfere constructively, giving a bright diffraction spot.

Bragg’s law is the foundation of X-ray crystallography — the technique that revealed the structure of DNA, proteins, and virtually every crystal structure we know.

Alternative Method

You can also use the reciprocal lattice approach: the diffraction condition becomes $\vec{G} = \vec{k'} - \vec{k}$ , where $\vec{G}$ is a reciprocal lattice vector. This is the Laue condition, equivalent to Bragg’s law. For JEE, Bragg’s law is sufficient.

The angle $\theta$ in Bragg’s law is the glancing angle (angle with the plane), not the angle with the normal. This is different from the convention in optics (Snell’s law uses angle with normal). If $\theta$ is the glancing angle, the angle with the normal is $90° - \theta$ . Watch the question carefully for which angle is given.

Common Mistake

Students often confuse the glancing angle with the angle of incidence. In Bragg’s law, $\theta$ is measured from the crystal plane, not from the normal. If a question gives the angle of incidence (from the normal) as $\alpha$ , then $\theta = 90° - \alpha$ . Substituting $\alpha$ directly into $2d\sin\theta$ gives wrong results.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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