Question

Monochromatic X-rays having energy $2.8 \times 10^{-15}$ J diffracted (first order) from (200) plane of a cubic crystal at an angle 8.5°. The length of unit cell in of the crystal (rounded off to one decimal place) is 
(Given:} Planck’s constant, $h = 6.626 \times 10^{-34}$ J·s; $c = 3.0 \times 10^8$ m/s{)

• A. 2.4
• B. 3.4
• C. 4.8
• D. 9.8

Hint:

Use $E = hc/\lambda$ to find wavelength from X-ray energy, and then apply Bragg’s law $n\lambda = 2d\sin\theta$ to find lattice spacing or unit cell length.

Answer 1

The Correct Option is C

Step 1: Determine wavelength from given energy.
\[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \] \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2.8 \times 10^{-15}} = 7.1 \times 10^{-11} \text{ m} = 0.71 \text{ Å} \] Step 2: Apply Bragg’s equation.
\[ n\lambda = 2d\sin\theta \] For first order ($n=1$): \[ d = \frac{\lambda}{2\sin\theta} = \frac{0.71}{2\sin8.5^\circ} = \frac{0.71}{0.296} \approx 2.4 \text{ Å} \] Step 3: Relate interplanar distance to unit cell length.
For cubic crystal and (200) plane, \[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} = \frac{a}{\sqrt{8}} = \frac{a}{2} \] \[ a = 2d = 2 \times 2.4 = 4.8 \text{ Å} \] Step 4: Conclusion.
Hence, the unit cell length is 4.8 Å.