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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.A. `Z = 6`B. `Z = 4`C. `Z = 11`D. `Z = 44` |
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Answer» Correct Answer - A `sqrt(f_(1)) = sqrt((v)/(lambda_(1))) = a(11 - 1)` and `sqrt(f_(2)) = sqrt((v)/(lambda_(2))) = a(Z - 1)` "By dividing" , `sqrt((lambda_(1))/(lambda_(2))) = (10)/(Z - 1) rArr sqrt((4)/(1)) = (10)/(Z - 1) rArr Z = 6` |
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