Question

For a simple cubic crystal, the smallest inter-planar spacing d that can be determined from its second order of diffraction using monochromatic X-rays of wavelength 1.32Å is _________Å. (Round off to two decimal places)

Answer 1

Correct Answer: 1.32

Bragg’s Law and Inter-Planar Spacing 

For a simple cubic crystal, the inter-planar spacing \( d \) for the \( n \)-th order diffraction is given by Bragg’s Law:

\[ n\lambda = 2d \sin\theta \]

Given Data

• Order of diffraction: \( n = 2 \) (second-order diffraction)
• Wavelength of X-rays: \( \lambda = 1.32 \) Å
• Angle of diffraction: \( \theta \)

Step 1: Finding the Smallest Inter-Planar Spacing

The smallest inter-planar spacing occurs when \( \theta = 90^\circ \), which gives \( \sin\theta = 1 \).

Thus, the equation simplifies to:

\[ 2\lambda = 2d \]

Step 2: Solving for \( d \)

Rearranging the equation:

\[ d = \lambda \]

Substituting \( \lambda = 1.32 \) Å:

\[ d = 1.32 \text{ Å} \]

Final Answer

Thus, the smallest inter-planar spacing is \( 1.32 \) Å.