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Q.
If the wavelength of first member of Balmer series in hydrogen spectrum is $6563\,\mathring{A}$ , the wavelength of second member of Balmer series will be :
J & K CETJ & K CET 2001
Solution:
Balmer found that the wavelength of all lines in hydrogen spectrum can be represented by
$\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$
where $R$ is Rydbergs constant.
Given, $n_{2}=3, n_{1}=2$
$\therefore \frac{1}{\lambda_{1}}=R\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 R}{36}$
$\Rightarrow \frac{1}{\lambda_{2}}=R\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 R}{16} $
$\Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{16}{3 R} \times \frac{5 R}{36} $
$\Rightarrow \lambda_{2}=\frac{80}{108} \lambda_{1}$
Given, $ \lambda_{1}=6563\,\mathring{A}$
$\therefore \lambda_{2}=\frac{80}{180} \times 6563=4861\,\mathring{A}$