Let ${\lambda }_{\alpha }$, ${\lambda }_{\beta }$ and ${\lambda }_{\alpha }^{\prime }$ denote the wavelengths of the X-rays of the ${\text{K}}_{\alpha }$, ${\text{K}}_{\beta }$ and ${\text{L}}_{\alpha }$ lines in the characteristic X-rays for a metal. Then,

• Correct answer
  A

  $\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{\prime }}$
• B

  $\frac{1}{\lambda {\text{'}}_{\alpha }}=\frac{1}{{\lambda }_{\beta }}+\frac{1}{{\lambda }_{\alpha }}$
• C

  $\frac{1}{{\lambda }_{\alpha }}=\frac{1}{{\lambda }_{\beta }}+\frac{1}{\lambda {\text{'}}_{\alpha }}$
• D

  ${\lambda }_{\alpha }={\lambda }_{\beta }+\lambda {\text{'}}_{\alpha }$

Solution

${\text{E}}_{\text{L}}-{\text{E}}_{\text{K}}=\frac{\text{hc}}{{\text{λ}}_{\text{α}}}$ ...(i)


${\text{E}}_{\text{M}}-{\text{E}}_{\text{K}}=\frac{\text{hc}}{{\lambda }_{\beta }}$ ...(ii)

${\text{E}}_{\text{M}}-{\text{E}}_{\text{L}}=\frac{\text{hc}}{{\lambda }_{\alpha }^{\text{'}}}$ ...(iii)

$\left(\text{ii}\right)-\left(\text{i}\right)\Rightarrow {\text{E}}_{\text{M}}-{\text{E}}_{\text{L}}=\frac{\text{hc}}{{\lambda }_{\alpha }^{\text{'}}}=\frac{\text{hc}}{{\lambda }_{\beta }}-\frac{\text{hc}}{{\lambda }_{\alpha }}$

$\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{\text{'}}}$
 

 

Hence, the correct answer is option: 'A'.