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

The Correct Option is B

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