X-rays of wavelength 0.103 nm reflects off a crystal and a second-order maximum is recorded at a Bragg angle of \(25.5^{\circ}\). What is the spacing between the scattering planes in this crystal?

Short Answer

Expert verified
The spacing between the scattering planes in the crystal is approximately 0.182 nm.

Step by step solution

01

List down the given parameters

We are given: - Wavelength (\(\lambda\)) = 0.103 nm - Bragg angle (\(\theta\)) = 25.5° - Order of maximum reflection (n) = 2
02

Convert the Bragg angle to radians

To work with trigonometric functions in calculations, we need to convert the angle from degrees to radians. The conversion between degrees and radians is \(1 radian = \frac{180}{\pi}°\). Bragg angle in radians: \(\theta = 25.5° \times \frac{\pi}{180} \approx 0.445 rad\)
03

Apply Bragg's Law to find the spacing between the scattering planes

Now we have all the required variables to calculate the spacing, we will use Bragg's Law: \(n\lambda = 2d\sin{\theta}\) and rearrange the equation to solve for \(d\): \(d = \frac{n\lambda}{2\sin{\theta}}\)
04

Calculate the spacing between the scattering planes

Substitute the given values into the equation: \(d = \frac{(2)(0.103 nm)}{2\sin{(0.445)}}\) Now, calculate the value of \(d\): \(d \approx 0.182 nm\)
05

Report the result

The spacing between the scattering planes in the crystal is approximately 0.182 nm.

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