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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.

Text Solution

Verified by Experts

`lambda_(1)/lambda_(2)=(Z_(2)-1)^(2)/(Z_(1)-1)^(2)(since,1/lambdaalpha(Z-1)^(2))`
` 1/4=(Z_(2)-1)^(2)/(11-1)^(2)`on solving `Z_(2)=6`
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