The element which has a $${K_\alpha }$$ x-rays line of wavelength $$1.8\,\mathop {\text{A}}\limits^ \circ $$  is ($$R = 1.1 \times {10^{ - 7}}{m^{ - 1}},b = 1$$      and $$\sqrt {\frac{5}{{33}}} = 0.39$$   )

Solution :
$$\eqalign{ & \frac{1}{{{\lambda _\alpha }}} = \frac{{3R}}{4}{\left( {Z - 1} \right)^2} \cr & \Rightarrow \left( {Z - 1} \right) = \sqrt {\frac{4}{{3 \times 1.1 \times {{10}^7} \times 1.8 \times {{10}^{ - 10}}}}} \cr & = \frac{{200}}{3}\sqrt {\frac{5}{{33}}} = \frac{{78}}{3} = 26 \cr & \Rightarrow Z = 26 + 1 = 27 \cr} $$