Question

When the potential difference applied to an x-ray tube is $10\text{kV}$ , the minimum wavelength of continuous x-ray spectrum is ${\lambda }_0.$ Further when the potential difference applied is increased from $10\text{kV}\text{to}20\text{kV}$ , the difference of wavelengths of $K_0$ line and minimum wavelength of continuous x- ray increases to twice of its initial value . Atomic number of anode material of x-ray tube is $Z$. If the value of $Z-1$ equals $\sqrt{\frac{n}{g{\lambda }_{0}R}},$ then find the value of $n$ ______.

Answer 1

Given,

${\lambda }_{k_{\alpha }}-{\lambda }_0^{\prime }=2({\lambda }_{k_{\alpha }}-{\lambda }_0)$

$\Rightarrow {\lambda }_{k_{\alpha }}=2{\lambda }_0-{\lambda }_0^{\prime }$

$\frac{{\lambda }_0^{\prime }}{{\lambda }_0}=\frac{10}{20}=\frac{1}{2}[\because {\lambda }_{min}=\frac{hc}{eV}]$

$\Rightarrow {\lambda }_{k_{\alpha }}=2{\lambda }_0-\frac{{\lambda }_0}{2}=\frac{3{\lambda }_0}{2}\Rightarrow \frac{4}{3R(Z-1{)}^2}=\frac{3{\lambda }_0}{2}\Rightarrow Z-1=\sqrt{\frac{8}{9{\lambda }_{0}R}}$

Ans : 8