Question

The first member of the Balmer series of hydrogen atom has a wavelength of $6561\stackrel{\text{o}}{A}$. The wavelength of the second member of the Balmer series (in $\text{nm}$) is

Answer 1

The wavelength of the spectral line of hydrogen spectrum is given by the formula
$\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
Where,$R=$Rydberg constant
For the first member of Balmer series $n_f=2,n_i=3$
$\therefore \frac{1}{\lambda }=R(\frac{1}{2^2}-\frac{1}{3^2})...\left(i\right)$
For last member of Balmer series $n_f=2,n_i=4$
So, $\frac{1}{{\lambda }^{\prime }}=R[\frac{1}{4}-\frac{1}{16}]....\left(ii\right)$
Dividing $\left(i\right)$ by $\left(ii\right)$, we get
$\Rightarrow \frac{{\lambda }^{\prime }}{\lambda }=\frac{5\times 16}{9\times 4\times 3}$
$\Rightarrow {\lambda }^{\prime }=\frac{5\times 4\times 656.1}{9\times 3}\text{nm}=486\text{nm}$