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

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