Solid State

To find the distance between the layers in a crystalline solid that produces a diffraction peak at $2\theta = 60^{\circ}$, we first need to determine $\theta$. Given that $2\theta = 60^{\circ}$, it follows that $\theta = 30^{\circ}$.Then, we use Bragg's law for diffraction, which is:$n\lambda = 2d \sin\theta$Given:Wavelength of X-rays, $\lambda = 1.54 \text{\AA} = 1.54 \times 10^{-10} \text{m}$Order of reflection, $n=1$$\sin 30^{\circ} = 0.5$Rearranging Bragg's law to solve for $d$, we have:$d = \frac{n\lambda}{2\sin\theta}$Substituting the given values into the equation:$d = \frac{1 \times 1.54 \times 10^{-10}}{2 \times 0.5} = \frac{1.54 \times 10^{-10}}{1} = 1.54 \times 10^{-10} \text{m} = 1.54 \times 10^{-8} \text{cm}$Thus, the distance between the layers is $1.54 \times 10^{-8} \text{cm}$.