Answer
$17.46\;keV$
Work Step by Step
The wavelength corresponding to the $K_\alpha$ radiation is: $\lambda_\alpha=71.0\;pm$
Therefore, the photon energy corresponding to the $K_\alpha$ radiation is:
$E_\alpha=\frac{hc}{\lambda_\alpha}$
or, $E_\alpha=\frac{1240\;keV-pm}{\lambda_\alpha\;(pm)}$
or, $E_\alpha=\frac{1240\;keV-pm}{71.0\;pm}$
or, $E_\alpha=17.46\;keV$
