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Consider a silver target in coolidge tube to produce x-rays. The accelerating potential is 31 kV. `E_(K)=25.51 KeV, E_(L)=3.5 1 KeV. "If" lambda_(K alpha)-lambda_("min")` is approximately8N pm(in pm), where N is an integer find N. Round off to nearest integer. (Take : hc = 1240 e V nm) |
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Answer» Correct Answer - `0002` `lambda_("min")=(1240)/(31xx10^(3))=40xx10^(-2) nm = 4xx10^(-11)m=40 `pm `E_(k)-E_(L)=(hc)/(lambda_(k alpha))=25.51-3.5` `lambda_(k alpha)=(1240)/(22)xx10^(-3)=56.36` pm `=16.36 ~=16`pm |
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