InterviewSolution
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The interplaner distance in a crystal is `2.8 xx 10^(–8)` m. The value of maximum wavelength which can be diffracted : -A. `2.8 xx 10^(–8)` mB. `5.6 xx 10^(–8)` mC. `1.4 xx 10^(–8)` mD. `7.6 xx 10^(–8)` m |
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Answer» Correct Answer - B `2dsin theta=nlambda because -1 le sin theta le 1` Therefore `lambda_"max"=2d rArr lambda_"max"=2xx2.8xx10^(-8)m` `rArr lambda_"max"=5.6xx10^(-8)m` |
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