Question

What is the wavelength of the X-rays if first-order Bragg diffraction is observed at \(23.0^{\circ}\) relative to the crystal surface, with an interatomic distance of \(0.256 \mathrm{nm} ?\)

Short answer

Question: Given the first-order Bragg diffraction angle of 23.0 degrees and the interatomic distance of 0.256 nm, determine the wavelength of the X-rays. Answer: The wavelength of the X-rays is 0.2004 nm.

Step-by-step solution

Understand the given information and the required output

In this problem, we are given the first-order Bragg diffraction angle \(\theta = 23.0^\circ\) and the interatomic distance \(d = 0.256 \,\mathrm{nm}\). We need to find the wavelength of the X-rays, which we will denote as \(\lambda\).

Recall Bragg's Law and the relation between angle and order

Bragg's Law states that when X-rays are diffracted by a crystal lattice, the following condition must be satisfied: $$ n\lambda = 2 d \sin{\theta} $$ Here, \(n\) is the order of the diffraction, \(\lambda\) is the wavelength of the X-rays, \(d\) is the interatomic distance or lattice spacing, and \(\theta\) is the angle of incidence (diffraction angle). In this problem, we know the first-order diffraction, which means \(n = 1\).

Apply Bragg's Law to the given data

Using the values of \(n, d,\) and \(\theta\), we can plug them into Bragg's Law equation to find \(\lambda\): $$ \lambda = \frac{2 d \sin{\theta}}{n} $$ For the given data, we have: $$ \lambda = \frac{2(0.256) \sin{(23.0^\circ)}}{1} $$

Calculate the wavelength of the X-rays

After plugging in the values for \(d, \theta,\) and \(n\), we can calculate the wavelength \(\lambda\): $$ \lambda = \frac{2(0.256) \sin{(23.0^\circ)}}{1} = 2(0.256) \times 0.3919 = 0.2004\, \mathrm{nm} $$ So, the wavelength of the X-rays is \(0.2004\, \mathrm{nm}\).