`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.

Answer

6

Explanation

`lambda_(1)/(lambda_(2)) = ((Z_(2) - 1)^(2))/((Z_(1) - (1)^(2))) ((since, (1)/(lambda) prop (Z - 1) ^(2)))`
`(1)/(4) = ((Z_(2) - 1)^(2))/((11 - 1)^(2)) : "on solving" , Z_(2) = 6`