GATE 2025 PH – Question 33
Powder X-ray diffraction of a cubic solid with lattice constant $a$ has the (111) diffraction peak at $\theta=30^\circ$. If the lattice expands to $1.25a$, the angle corresponding to that peak changes to $\sin^{-1}(1/n)$. The value of $n$ (rounded off to one decimal place) is _____
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Correct answer: 2.49 to 2.51
Explanation
**Bragg's law:** $2d\sin\theta=\lambda$. For a cubic crystal $d_{111}=a/\sqrt3$, so $\sin\theta\propto1/a$ for a fixed wavelength.
**Original crystal:** $\theta=30^\circ$, so $\sin\theta=0.5$.
**Expanded lattice** $a^\prime=1.25a$:
$$\sin\theta^\prime=\frac{\sin\theta}{1.25}=\frac{0.5}{1.25}=0.4=\frac1{2.5}.$$
So $\theta^\prime=\sin^{-1}(1/n)$ with $n=\mathbf{2.5}$.