When the voltage applied to an X-ray tube increased from `V_(1)=15.5kV` to `V_(2)=31kV` the wavelength interval between the `K_(alpha)` line and the cut-off wavelength of te continuous X-ray spectrum increases by a factor of `1.3`. If te atomic number of the element of the target is z. Then the value of `(z)/(13)` will be: (take `hc=1240eVnm` and `R=1xx(10^(7))/(m))`

Answer

`26`

Explanation

`lambda_(th)=(hc)/(eV_(a))`
`1/(lambda_(Kalpha))=R(z-1)^(2)(1/1^(2)-1/2^(2))`
`13/10(lambda_(Kalpha-lambda_(th)))=(lambda_(Kalpha)-(lambda_(th))/2)`
`3/10 lambda_(Kalpha)=(13/10-1/2) lambda_(th)`
`3/10((4xx10^(-7))/(3(z_(7))^(2)))=(8/10)(12.4xx10^(-7))/(15.5xx10^(3)) Rightarrow 5000/8 = (z-1)^(2)`
`625=(z-1)^(2) Rightarrow z=26`