Question

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,

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

Answer 1

Answer: (A)

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

${\text{E}}_{\text{M}}-{\text{E}}_{\text{K}}=\frac{\text{hc}}{{\lambda }_{\beta }}$ ...(ii)
${\text{E}}_{\text{M}}-{\text{E}}_{\text{L}}=\frac{\text{hc}}{{\lambda }_{\alpha }^{{}^{\prime }}}$ ...(iii)
(ii) - (i) $\Rightarrow {\mathrm{E}}_{\mathrm{M}}-{\mathrm{E}}_{\mathrm{L}}=\frac{\mathrm{hc}}{{\lambda }_{\alpha }}=\frac{\mathrm{hc}}{{\lambda }_{\beta }}-\frac{\mathrm{hc}}{{\lambda }_{\alpha }}$
$\frac{1}{{\lambda }_{\beta }}=\frac{1}{{\lambda }_{\alpha }}+\frac{1}{{\lambda }_{\alpha }^{{}^{\prime }}}$