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: Recall Bragg's Law for X-ray diffraction. Bragg's Law is given by the formula: \[ n\lambda = 2d\sin\theta \] where \(n\) is the order of the maximum, \(\lambda\) is the wavelength of the X-rays, \(d\) is the distance between atomic planes, and \(\theta\) is the angle of incidence.
Step 2: Identify the given values from the problem. - Order of maximum, \(n = 1\) (since it's the "first maxima"). - Wavelength, \(\lambda = 0.32 \, \text{nm}\). - Angle of diffraction, \(\theta = 30^{\circ}\).
Step 3: Substitute the values into Bragg's Law and solve for \(d\). \[ (1)(0.32 \, \text{nm}) = 2 \cdot d \cdot \sin(30^{\circ}) \] We know that \(\sin(30^{\circ}) = 0.5\). \[ 0.32 \, \text{nm} = 2 \cdot d \cdot (0.5) \] \[ 0.32 \, \text{nm} = d \] Thus, the distance between the atomic planes is \(0.32 \, \text{nm}\).