\(\Rightarrow \,\frac{{{{({\lambda _{{K_\alpha }}})}_{Mo}}}}{{{{({\lambda _{{K_\alpha }}})}_{Zn}}}} = {\left( {\frac{{30 - 1}}{{42 - 1}}} \right)^2} = {\left( {\frac{{29}}{{41}}} \right)^2}\)
\(\therefore\) \(\,{({\lambda _{{K_a}}})_{Zn}} = \frac{{41}}{{29}} \times \frac{{41}}{{29}} \times 0.7078\,{\text{Å}} = {\text{1}}{\text{.414 Å}}\)