Question

The second-order diffraction $(n=2)$ for a gold crystal is at an angle of ${22.20}^{\circ }$ for $\mathrm{X}$ rays of $154\mathrm{pm}$. What is the spacing between these crystal planes?

Short answer

The spacing between the crystal planes in the gold crystal can be calculated using Bragg's Law formula: $n\lambda =2d\sin \theta$. Given the second-order diffraction (n=2), X-ray wavelength $\lambda =154\text{pm}$, and angle $\theta ={22.20}^{\circ }$, we can find the interplanar spacing (d) by rearranging the formula to $d=\frac{n\lambda }{2\sin \theta }$. Substituting the given values, we get $d\approx \frac{308\text{pm}}{\sin {22.20}^{\circ }}\approx 797.76\text{pm}$. Therefore, the spacing between the crystal planes is approximately $797.76\text{pm}$.

Step-by-step solution

Recall the Bragg's Law Formula

Bragg's Law states that for constructive interference to occur, the path difference between two rays reflecting off adjacent planes should be an integer multiple of the wavelength. The formula is given by: $n\lambda =2d\sin \theta$ where n = order of diffraction (integer), λ = wavelength of the incident ray, d = interplanar spacing (the distance between adjacent crystal planes), θ = angle of incidence/reflection.

Substitute the given values

We have been given the following data: n = 2 (second-order diffraction), λ = 154 pm, θ = 22.20 degrees. Now, put these values in Bragg's Law formula: $2(154\text{pm})=2d\sin {22.20}^{\circ }$

Solve for interplanar spacing (d)

Now, we need to find d by rearranging the formula and then solving it: $d=\frac{2(154\text{pm})}{2\sin {22.20}^{\circ }}$ Which simplifies to: $d=\frac{308\text{pm}}{\sin {22.20}^{\circ }}$ Now, find the value of d: $d=\frac{308\text{pm}}{\sin {22.20}^{\circ }}\approx 797.76\text{pm}$

Round the result and state the answer

After rounding the result to two decimal places, the spacing between these crystal planes is: $d\approx 797.76\text{pm}$ Thus, the spacing between the crystal planes in the gold crystal is approximately 797.76 pm.