The first member of the Balmer series of hydrogen atom has a wavelength of 6561 $\stackrel{0}{A}$. The wavelength of the second member of the Balmer series (in nm) is ----

$\begin{array}{l}Wavelength during transition can be given as,\text{}\\ \frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\\ For first member of Balmer Series of Hydrogen : \frac{1}{{\lambda }_1}=R{\left(1\right)}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)\\ For second member of Balmer Series of Hydrogen : \frac{1}{{\lambda }_2}=R{\left(1\right)}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3R}{16}\\ \therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{\raisebox{1ex}{$16$}\!\left/ \!\raisebox{-1ex}{$3R$}\right.}{\raisebox{1ex}{$36$}\!\left/ \!\raisebox{-1ex}{$5R$}\right.}=\frac{20}{27}\text{}\\ \Rightarrow {\lambda }_2=\frac{20}{27}\times 6561\stackrel{0}{A}=4860\stackrel{0}{A} =486nm\end{array}$