Question

The first maxima for Bragg's diffraction pattern by a crystal is observed at 30\(^{\circ}\) when X-rays wavelength of 0.32 nm are used. The distance between the atomic planes is:

• A. \( 0.32 \, \text{nm} \)
• B. \( 0.48 \, \text{nm} \)
• C. \( 0.84 \, \text{\AA} \)
• D. \( 0.48 \, \text{\AA} \)

Hint:

Bragg's Law is a fundamental equation in solid-state physics. Remember that \(n\) must be an integer (1, 2, 3, ...) representing the order of the reflection. For "first maxima" or "first-order diffraction," always use \(n=1\).

Answer 1

The Correct Option is A

Step 1: State Bragg's Law for X-ray diffraction. The formula is \[n\lambda = 2d\sin\theta\] where \(n\) represents the order of the maximum, \(\lambda\) is the X-ray wavelength, \(d\) is the interplanar spacing, and \(\theta\) is the angle of incidence.
Step 2: Extract the given parameters. These are \(n = 1\) (for the first maxima), \(\lambda = 0.32 \, \text{nm}\), and \(\theta = 30^{\circ}\).
Step 3: Solve for \(d\) by substituting the known values into Bragg's Law. \[(1)(0.32 \, \text{nm}) = 2 \cdot d \cdot \sin(30^{\circ})\] Given that \(\sin(30^{\circ}) = 0.5\), the equation becomes: \[0.32 \, \text{nm} = 2 \cdot d \cdot (0.5)\] Simplifying yields: \[0.32 \, \text{nm} = d\] Therefore, the distance between the atomic planes is \(0.32 \, \text{nm}\).