Verified Solution • 2025–26 Session moderate

QAIPM-2001-Ph-2932: NEET 2001 Physics - Class 12 AIPM-2001-Ph-2932 Solution: The interplanar distance in a crystal is \(2.8 10^{-8}\) m. The …

The interplanar distance in a crystal is 2.8×10−82.8\times10^{-8} m. The value of maximum wavelength which can be diffracted

Multiple Choice Options

  • A 2.8×10−82.8\times10^{-8} m
  • B 5.6×10−85.6\times10^{-8} m
  • C 1.4×10−81.4\times10^{-8} m
  • D 7.6×10−87.6\times10^{-8} m

Step-by-Step Verified Solution

2dsin _ = n _ _ _ 1 � sin _ _ � 1 Therefore _ max = 2d _ _ max = 2 _ 2.8 _ 10 _ 8 = 5.6 _ 10 _ 8 m

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