`k_(a) ` wavelength emitted by an atom of atomic number `Z= 11` is `lambda` find the atomic number for an atom that emils `k_(a)` radiation with wavelength `4 lambda`.

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`.