The wavelengths of the K_α and K_β X-rays of molybdenum are 0.71...

The energy of an X-ray photon is given by $E = \dfrac{hc}{\lambda}$.

The $K_{\alpha}$ transition corresponds to an electron jumping from the $L$ shell to the $K$ shell, so $E_{K_{\alpha}} = E_L - E_K$.

The $K_{\beta}$ transition corresponds to an electron jumping from the $M$ shell to the $K$ shell, so $E_{K_{\beta}} = E_M - E_K$.

The $L_{\alpha}$ transition corresponds to an electron jumping from the $M$ shell to the $L$ shell, so $E_{L_{\alpha}} = E_M - E_L$.

From these energy differences, we can write:

$E_{L_{\alpha}} = E_{K_{\beta}} - E_{K_{\alpha}}$

Substituting $E = \dfrac{hc}{\lambda}$:

$\dfrac{hc}{\lambda_{L_{\alpha}}} = \dfrac{hc}{\lambda_{K_{\beta}}} - \dfrac{hc}{\lambda_{K_{\alpha}}}$

$\dfrac{1}{\lambda_{L_{\alpha}}} = \dfrac{1}{\lambda_{K_{\beta}}} - \dfrac{1}{\lambda_{K_{\alpha}}}$

$\lambda_{L_{\alpha}} = \dfrac{\lambda_{K_{\alpha}} \lambda_{K_{\beta}}}{\lambda_{K_{\alpha}} - \lambda_{K_{\beta}}}$

Given $\lambda_{K_{\alpha}} = 0.71 \text{ \AA}$ and $\lambda_{K_{\beta}} = 0.63 \text{ \AA}$:

$\lambda_{L_{\alpha}} = \dfrac{0.71 \times 0.63}{0.71 - 0.63}$

$\lambda_{L_{\alpha}} = \dfrac{0.4473}{0.08} = 5.59125 \text{ \AA}$

Rounding to two decimal places, we get $5.59 \text{ \AA}$.