`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`.

Options

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

Answer

C, D

Explanation

According to Moseley's law,
`sqrt f = a (z - b) implies f = a^(2) (z- b)^(2)`
`implies ( c)/(lambda) = a^(2) (z- b)^(2)`
`For K_(a) line, b = 1`
`From (i) , (lambda_(2))/(lambda_(1)) = ((z_(1) - 1)^(2))/((z_(2) - 1)^(2)) implies (4 lambda)/(lambda) = ((11 - 1)^(2))/(z_(2) - 1)^(2)`
`implies Z_(2) - 1 = (10)/(2) implies z_(2) = 6`