Question

The interplaner distance in a crystal is 2.8 × 10-8 m. The value of maximum wavelength which can be diffracted : -

• A. 2.8 × 10-8 m
• B. 5.6 × 10-8 m
• C. 1.4 × 10-8 m
• D. 7.6 × 10-8 m

Answer 1

Answer: (B)

5.6 × 10-8 m

Answer 2

The correct option is (B): 5.6 × 10-8 m.
2d sin $\theta$ = n$\lambda$ $\because$ – 1 $\leq$ sin $\theta$ $\leq$ 1
Therefore $\lambda_{max}$ = 2d$\Rightarrow$ $\lambda_{max}$. = 2 × 2.8 × 10-8 m
\Rightarrow$$\lambda_{max}. = 5.6 × 10-8 m