Question

Let $\lambda _{\alpha }$ , $\lambda _{\beta }$ and $\lambda _{\alpha }^{'}$ 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. $\frac{1}{\lambda _{\beta }} = \frac{1}{\lambda _{\alpha }} + \frac{1}{\lambda _{\alpha }^{'}}$
• B. $\frac{1}{\lambda '_{\alpha }}=\frac{1}{\lambda _{\beta }}+\frac{1}{\lambda _{\alpha }}$
• C. $\frac{1}{\lambda _{\alpha }}=\frac{1}{\lambda _{\beta }}+\frac{1}{\lambda '_{\alpha }}$
• D. $\lambda _{\alpha }=\lambda _{\beta }+\lambda '_{\alpha }$

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