Question

The wavelength of $K_{\alpha }$ and $L_{\alpha }$ X-Rays of a material are 20 pm and 140 pm respectively. Find the wavelength of $K_{\beta }$ X-Ray of the material.

• A. 12.5 pm
• B. 15.5 pm
• C. 17.5 pm
• D. 19.5 pm

Answer 1

The correct option is C 17.5 pm

For the wavelength of $K_{\alpha }$ x-rays; $\frac{hc}{{\lambda }_1}$ = $E_K-E_L$
for the wavelength of $L_{\alpha }$ x-rays; $\frac{hc}{{\lambda }_2}$ = $E_L-E_M$
or the wavelength of $K_{\beta }$ x-rays; $\frac{hc}{{\lambda }_3}$ = $E_K-E_M$
Now,
If we add the first two equation we get :
$hc(\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2})=E_K-E_M$
Now from equations iii) and iv)
$\frac{hc}{{\lambda }_3}=hc(\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}).$
We get;
$\frac{1}{{\lambda }_3}=(\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}).$
Or,
$\frac{1}{{\lambda }_3}=(\frac{1}{20}+\frac{1}{140})p{m}^{-1}$
Or,
${\lambda }_3=\frac{2800}{160}pm.$
Or,
${\lambda }_3$ = 17.5 pm