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Q.
The colour of the second line of Balmer series is:
J & K CETJ & K CET 2004
Solution:
When an atom comes down from some higher energy level to the second energy $(n=2)$,
then the lines of spectrum are obtained in visible part and give the Balmer series. $\frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right),$
$n=3,4,5...$ For second line $n=4$
$\therefore \frac{1}{\lambda}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)$
$=\frac{3 R}{16}$ $\Rightarrow \lambda=\frac{16}{3 R}$
$R=1.097 \times 10^{7} m ^{-1}$
$\lambda=\frac{16}{3 \times 1.097 \times 10^{7}}$
$=4860 \times 10^{-10} m$
$\Rightarrow \lambda=4860\,\mathring{A}$ which corresponds to colour blue.