Answer
$\boxed{\lambda_{\alpha}=21.36\;pm}$
Work Step by Step
The $K_{\alpha}$ spectral line originates when an electron from the $L$ shell fills a hole in the $K$ shell.
The energy difference between $K$ and $L$ shells of tungsten is
$\Delta E=(69.5-11.3)\;keV=58.2\;keV$
Therefore, the wavelength of the $K_{\alpha}$ line for tungsten is
$\lambda_{\alpha}=\frac{hc}{\Delta E}=\frac{6.63\times 10^{-34}\times3\times3^{8}}{58.2\times10^3\times1.6\times 10^{-19}}\;m$
or, $\lambda_{\alpha}\approx2.136\times10^{-11}\;m$
or, $\boxed{\lambda_{\alpha}=21.36\;pm}$
