The glancing angle in a X-rays diffraction experiment is 30^° and the wavelength of the X rays used is 20 ~mm . The interplanar spacing of the crystal diffracting these X-rays will be

Correct answer: (B) 20 ~nm. Using formula 2 ~d sin θ=n λ d=(n λ)/(2 sin θ)=frac1 × 20 × 10^-32 sin 30^° d=frac20 × 10^-32 × 1/2=20 ~nm

Q98 / 110 · VITEEE 2022 · Physics

The glancing angle in a X-rays diffraction experiment is \(30^{\circ}\) and the wavelength of the \(\mathrm{X}\) rays used is \(20 \mathrm{~mm}\). The interplanar spacing of the crystal diffracting these X-rays will be

A
\(40 \mathrm{~nm}\)
Correct answer
B
\(20 \mathrm{~nm}\)
C
\(15 \mathrm{~nm}\)
D
\(10 \mathrm{~nm}\)

Solution

Using formula \(\begin{aligned} & 2 \mathrm{~d} \sin \theta=\mathrm{n} \lambda \\ & \mathrm{d}=\frac{\mathrm{n} \lambda}{2 \sin \theta}=\frac{1 \times 20 \times 10^{-3}}{2 \sin 30^{\circ}} \end{aligned}\) \(\mathrm{d}=\frac{20 \times 10^{-3}}{2 \times \frac{1}{2}}=20 \mathrm{~nm}\)

Hence, the correct answer is option: 'B'.

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