`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. 6
B. 4
C. 11
D. 44

Answer

A

Explanation

From Moseley's law , `sqrt v = a (Z - 1)` for `K_(alpha)` X-ray
i.e. , `(1)/(sqrt lambda) = a (Z - 1)`
Given, `(1)/(sqrt lambda) = a (Z - 1)` and `(1)/(sqrt 4 lambda) = a (Z - 1)`
Driving we get `Z = 6`.