The interplaner distance in a crystal is 2.8 × 10-8 m. The value of maximum wavelength which can be diffracted : -
The interplaner distance in a crystal is 2.8 × 10-8 m. The value of maximum wavelength which can be diffracted : -
2.8 × 10-8 m
5.6 × 10-8 m
1.4 × 10-8 m
7.6 × 10-8 m
The Correct Option is B
Solution and Explanation
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
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Concepts Used:
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In the single-slit diffraction experiment, we can examine the bending phenomenon of light or diffraction that causes light from a coherent source to hinder itself and produce an extraordinary pattern on the screen called the diffraction pattern.
Read More: Difference Between Diffraction and Interference
Central Maximum
