Question

The wavelength of the first line in balmer series in the hydrogen spectrum is $\lambda$. The wavelength of the second line is $\frac{y\lambda }{27}$. Find $y$.

Answer 1

Answer: 20

$\frac{1}{{\lambda }_1}=R\left(\frac{1}{4}-\frac{1}{9}\right)$
$\Rightarrow {\lambda }_1=\frac{4\times 9}{5R}$
or, $\lambda =\frac{4\times 9}{5R}\left(\because {\lambda }_1=\lambda \right)$
$\frac{1}{{\lambda }_1}=R(\frac{1}{4}$ or, $\lambda =\frac{4\times }{5R}$
Similarly,
$\frac{1}{{\lambda }_2}=R\left(\frac{1}{4}-\frac{1}{4^2}\right)$
$\Rightarrow {\lambda }_2=\frac{16}{3R}=\frac{16}{3}\times \frac{5\lambda }{4\times 9}=\frac{20}{27}\lambda$