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Question

When another X-rays source was used, a reflection was observed at 1526. The wavelength of the X-rays was (in pm)?

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Solution

Firstcase:λ=229pmQ=23020Interplanarspacingd=λ2sinQd=229210.396d=289.14pmSecondcase:d=289.14pmQ=15026λ=?λ=2dsinQ=2×(289.14)(sin(15026))=2×1289.14(0.266)λ=153.82pm

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