Question

The diffraction pattern of a crystalline solid gave a peak at \( 2\theta = 60^\circ \). What is the distance (in cm) between the layers that gave this peak? \[ \text{(Given: Wavelength } \lambda = 1.544\text{ Å, } \sin 30^\circ = 0.5, \sin 60^\circ = 0.866, n=1) \]

• A. \( 8.89 \times 10^{-9} \) cm
• B. \( 8.89 \times 10^{-1} \) cm
• C. \( 1.54 \times 10^{-8} \) cm
• D. \( 1.54 \) cm

Hint:

Bragg’s law relates the wavelength, diffraction angle, and interplanar spacing in a crystal lattice.

Answer 1

The Correct Option is C

Step 1: Using Bragg’s Law. Bragg’s equation: \[ n\lambda = 2d \sin\theta \] For first-order diffraction (\( n=1 \)): \[ d = \frac{\lambda}{2\sin\theta} \] Step 2: Substituting Values. \[ d = \frac{1.544 \times 10^{-8} \text{ cm}}{2 \times 0.866} \] \[ = \frac{1.544 \times 10^{-8}}{1.732} \] \[ \approx 1.54 \times 10^{-8} \text{ cm} \] Thus, the correct answer is (3).