If there times of `lambda_(min)` of continuous `X-`ray spectrum of target metal at `40 kV` is some as the wavelength of `K_(alpha)` line of this metal at `30 kV`, then determine the atomic number of the target metal.

Answer

`37`

Explanation

`:' lambda_(min)=(hc)/(eV)=12400/V Å`
At `40 kV : lambda_(min)=12400/40000=0.31 Å`
Wavelength of `K_(alpha)` is independent of applied potential.
For `K_(alpha)` X-ray : `3/4 (13.6)(Z-1)^(2)=E=(hc)/(lambda_(K alpha))`
`alpha_(K_(alpha))=(1216)/((Z-1)^(2)) Å` and given that
`lambda_(K_(alpha))=3 lambda_(min)implies (1216)/((Z-1)^(2))=3xx0.31`
`implies (Z-1)^(2)=1216/0.93 cong 1308 implies Z-1=36implies Z=37`