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Atoms and Nuclei question

2001 · Q143
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Atoms and Nuclei question

2001 · Q143

NEETPhysicsAtoms and NucleiMCQ+4 / −1
The interplanar distance in a crystal is 2.8 ×\times× 10−-−8 m. The value of maximum wavelength which can be diffracted
  1. A
    2.8 ×\times× 10−-−8 m
  2. B
    5.6 ×\times× 10−-−8 m
  3. C
    1.4 ×\times× 10−-−8 m
  4. D
    7.6 ×\times× 10−-−8 m
View written solutionFree

Correct answer: B

2dsinθ\theta θ = nλ\lambda λ

∵\because∵ −1≤sin⁡θ≤1 - 1 \le \sin \theta \le 1−1≤sinθ≤1

Therefore λ\lambda λmax = 2d

⇒\Rightarrow⇒ λ\lambda λmax = 2 ×\times× 2.8 ×\times× 10−-−8

= 5.6 ×\times× 10−-−8 m

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