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If the wavelength of $1^{\text {st }}$ line of Balmer series of hydrogen is $6561 Å$, the wavelength of the $2^{\text {nd }}$ line of series will be
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$4860 Å$
For the first line of Balmer series,
$\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}$
For the second line of Balmer series.
$\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16}$
$\therefore \frac{\lambda_2}{\lambda_1}=\frac{5 R / 36}{3 R / 16}=\frac{20}{27}$;
or $\lambda_2=\frac{20}{27}(6561 Å)=4860 Å$
$\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}$
For the second line of Balmer series.
$\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16}$
$\therefore \frac{\lambda_2}{\lambda_1}=\frac{5 R / 36}{3 R / 16}=\frac{20}{27}$;
or $\lambda_2=\frac{20}{27}(6561 Å)=4860 Å$
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