Question

In an X-ray diffraction experiment with Cu crystals having lattice parameter \( 3.61 \, \text{Å} \), X-rays of wavelength of \( 0.090 \, \text{nm} \) are incident on the family of planes \( \{ 1 1 0 \} \). The highest order present in the diffraction pattern is .......

Hint:

In X-ray diffraction, the highest diffraction order can be calculated using Bragg's law and the distance between the planes.

Answer 1

Correct Answer: 5

Step 1: Use the Bragg's law.
Bragg’s law for diffraction is given by \[ n \lambda = 2 d \sin \theta, \] where \( n \) is the diffraction order, \( \lambda \) is the wavelength, \( d \) is the distance between the planes, and \( \theta \) is the diffraction angle. Step 2: Calculate the distance between planes.
The lattice parameter \( a = 3.61 \, \text{Å} \) and for the \( \{ 1 1 0 \} \) planes, \[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} = \frac{3.61}{\sqrt{1^2 + 1^2 + 0^2}} = \frac{3.61}{\sqrt{2}} = 2.55 \, \text{Å}. \] Step 3: Apply Bragg’s law for maximum diffraction order.
For the highest order diffraction, \( \theta = 90^\circ \), so \[ n = \frac{2 d}{\lambda}. \] Substitute the values \( d = 2.55 \, \text{Å} \) and \( \lambda = 0.090 \, \text{nm} = 0.090 \times 10^{-1} \, \text{Å} \): \[ n = \frac{2 \times 2.55}{0.090} \approx 56.67. \] Thus, the highest integer \( n \) is 56. Final Answer: The highest order present in the diffraction pattern is \( \boxed{56}. \)