Question

When the voltage applied to an X-ray tube increases from $V_1=10kV$ to $V_2=20kV$, the wavelength interval between $K_a$ line and cut-off wavelength of continuous spectrum increases by a factor of $3$. Find out the Atomic number of the metallic target.

Answer 1

$\frac{1}{{\lambda }_L}=R{(z-1)}^2(\frac{1}{1^2}-\frac{1}{z^2})$

$=\frac{3}{4}R{(z-1)}^2$

$\Rightarrow {\lambda }_L=\frac{4}{3R{(z-1)}^2}$

for wttoff wave length :

$\Rightarrow {\lambda }_1=\frac{h_1}{e{v}_1}$ $\&$ ${\lambda }_2=\frac{h_1}{e{v}_2}$

$\Rightarrow ({\lambda }_2-{\lambda }_1)=3({\lambda }_1-{\lambda }_2)$

$\Rightarrow 3{\lambda }_1-{\lambda }_2=2{\lambda }_2$

$\Rightarrow \frac{h_L}{e}(\frac{3}{v_1}-\frac{1}{v_2})=8\left(\frac{1}{3R{(z-1)}^2}\right)$

$\Rightarrow v_1=10kv,$ $v_2=20kv$

putting value

$\Rightarrow z=29$

Hence, the answer is $29.$