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NEET 2001 (AIPMT) · Quantum Numbers, X-rays & Other Questions

Quantum Numbers, X-rays & Other Questions question from NEET 2001 (AIPMT)

Q9NEET 2001 · AIPMTMedium

Quantum Numbers, X-rays & Other Questions

The interplanar distance in a crystal is 2.8 ×\times 10−-8 m. The value of maximum wavelength which can be diffracted

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Answer: (b)

  1. Bragg's law: 2dsin⁡θ=nλ2d\sin\theta = n\lambda
  2. Maximum λ\lambda when sin⁡θ=1\sin\theta = 1, n=1n = 1: λmax=2d\lambda_{max} = 2d
  3. λmax=2×2.8×10−8=5.6×10−8\lambda_{max} = 2\times 2.8\times 10^{-8} = 5.6\times 10^{-8} m
  4. So the answer is (b) 5.6×10−85.6 \times 10^{-8} m.
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