Calculating X-ray Wavelength for BCC Iron Diffraction

To find the X-ray wavelength ($\lambda$), we use Bragg's Law for diffraction:

$ \lambda = \frac{2d \sin\theta}{n} $

Here, $n$ is the order of diffraction (for the first peak, $n=1$), $d$ is the interplanar spacing, and $\theta$ is the Bragg angle.

Determining Bragg Angle ($\theta$)

The given diffraction angle is $2\theta = 68.7^\circ$. The Bragg angle $\theta$ is half of this value:

$ \theta = \frac{2\theta}{2} = \frac{68.7^\circ}{2} = 34.35^\circ $

Finding Interplanar Spacing ($d$) for BCC

For a Body-Centered Cubic (BCC) crystal structure, the allowed reflections occur when the sum of Miller indices ($h+k+l$) is even. The first diffraction peak typically corresponds to the plane with the smallest value of $h^2 + k^2 + l^2$ that satisfies this condition, which is the (110) plane.

The formula for interplanar spacing ($d_{hkl}$) is:

$ d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} $

For the (110) plane in BCC iron:

$ d_{110} = \frac{a}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{a}{\sqrt{2}} $

Given the lattice parameter $a = 0.287 \text{ nm}$:

$ d_{110} = \frac{0.287 \text{ nm}}{\sqrt{2}} \approx 0.20294 \text{ nm} $

Calculating X-ray Wavelength ($\lambda$)

Now, substitute the values into Bragg's Law (with $n=1$):

$ \lambda = 2 d_{110} \sin\theta $

$ \lambda = 2 \times (0.20294 \text{ nm}) \times \sin(34.35^\circ) $

Using $\sin(34.35^\circ) \approx 0.56476$:

$ \lambda \approx 2 \times 0.20294 \text{ nm} \times 0.56476 $

$ \lambda \approx 0.22914 \text{ nm} $

Rounding to three decimal places, the wavelength is $0.229 \text{ nm}$.