Question

The $K_{\alpha }$ and $K_{\beta }\text{X}-\text{rays}$ of molybdenum have wavelengths $0.7\dot{\text{A}}$ and $0.6\dot{\text{A}}$ respectively. The wavelength $\left(\text{in}\dot{\text{A}}\right)$ of $L_{\alpha }\text{X}$- rays of molybdenum

Answer 1

Given:
${\lambda }_{K_{\alpha }}=0.7\dot{\text{A}};{\lambda }_{K_{\beta }}=0.6\dot{\text{A}}$


From diagram,

$E_3=E_1+E_2$

As we know, $E=\frac{hc}{\lambda }$

$\frac{hc}{{\lambda }_{K_{\beta }}}=\frac{hc}{{\lambda }_{K_{\alpha }}}+\frac{hc}{{\lambda }_{L_{\alpha }}}$

$\Rightarrow {\lambda }_{L_{\alpha }}=\frac{{\lambda }_{K_{\beta }}\times {\lambda }_{K_{\alpha }}}{{\lambda }_{K_{\alpha }}-{\lambda }_{K_{\beta }}}=\frac{0.6\times 0.7}{0.7-0.6}$

$\therefore {\lambda }_{L_{\alpha }}=0.42\dot{\text{A}}$