X-rays emitted from a copper target and a molybdenum target are found to contains a line of wavelength `22.85nm` attributed to the `K_(alpha)` line of an impurity element. The `K_(alpha)` lines of a copper `(Z=29)` and molybdenum `(Z=42)` have wavelength `15.42nm` and `7.12nm` respectively. Using Moseley's law, `gamma^(1//2)=a(Z-b)`. Calculate the atomic number of the impurity element.

Answer

24

Explanation

` sqrtV=sqrt((hc)/(lambda))=a(z-b)`
for copper `sqrt((hc)/(15.42xx10^(-9)))=a(29-b)" "..........(i)`
for molybdenum `sqrt((hc)/(7.12xx10^(-9)))=a(42-b)" "..........(ii)`
Finding the value of a & b and putting it for impurity element, we get
`z=24`