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`K_alpha` wavelength emitted by an atom of atomic number Z=11 is `lambda`. Find the atomic number for an atom that emits `K_alpha` radiation with wavelength `4lambda`. (a) Z=6 (b) Z=4 (c ) Z=11 (d) Z=44. |
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Answer» Correct Answer - A `(1)/(lambda) prop (Z-1)^2` `:. (lambda_1)/(lambda_2) = ((Z_2- 1)/(Z_1 - 1))^2 or (1)/(4) = ((Z_2 -1)/(11-1))^2` Solving this , we get Z_2 = 6 `:.` Correct answer is (a). |
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