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Q.
What is the wave number of $4^{th}$ line in Balmer series of hydrogen spectrum? $(R=1,09,677\,cm^{-1})$
Punjab PMETPunjab PMET 2008Structure of Atom
Solution:
$\bar{v}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
For Balmer series, $n_{1}=2$ and for $4^{\text {th }}$ line in Balmer series $n_{2}=6$
$\because R =109677\, cm ^{-1}$
$\bar{v} =109677\left(\frac{1}{2^{2}}-\frac{1}{6^{2}}\right)$
$=109677\left(\frac{1}{4}-\frac{1}{36}\right)$
$\bar{v} =24,372\, cm ^{-1}$