Question

When the voltage applied to an x-ray tube is increased from 10 kV to 20kV, the wavelength interval between the $K_{\alpha }$ line and the short wave cut off of the continuous $X$-ray spectrum increases by a factor $3$. Find the atomic number of the element of which the tube anti cathode is made. (Rydberg's constant = ${10}^7{m}^{-1}$).

Answer 1

At $10kV$, let the wavelength of $K_{\alpha }$ be ${\lambda }_1$ and the that of short wave be ${\lambda }_2$.

So ${\lambda }_2=hc/e\times 10kV=1.2375\dot{A}$.

Now at $20kV$, there is no effect on wavelength of $K_{\alpha }$ line as it depends only on the material properties. But the wavelength of the short wave gets halved to ${\lambda }_2/2$.

Now according to the question,

${\lambda }_1-{\lambda }_2/2=3({\lambda }_1-{\lambda }_2)\Rightarrow {\lambda }_1=5{\lambda }_2/4=1.547\dot{A}$

As the wavelength of $K_{\alpha }$ line is given as ,

$1/{\lambda }_1=(Z-1{)}^{2}R(1/1^2-1/2^2)=(Z-1{)}^{2}R\times \frac{3}{4}$

$\Rightarrow (Z-1{)}^2\approx 862$

$\Rightarrow Z=30.36\approx 30$

where $Z$ is the atomic number of the element.