`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.

Options

A. `Z = 6`
B. `Z = 4`
C. `Z = 11`
D. `Z = 44`

Answer

A

Explanation

`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`