If the wavelength of the first line in Balmer series is 656 nm, then the wavelength of its second line and…
If the wavelength of the first line in Balmer series is 656 nm, then the wavelength of its second line and limiting line respectively are
485.9 nm and 434 nm
485.9 nm and 364.4 nm
715 nm and 434 nm
608 nm and 415.2 nm
Solution
According to H-spectrum;
Rydberg formula $\Rightarrow \frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}ight)$
For first line, $n_1=2, n_2=3$ (Balmer series)
$\lambda=656 \mathrm{~nm}$
$\begin{aligned} \frac{1}{656} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}ight] \\ & =\frac{5}{36} R_{\mathrm{H}}...(i)\end{aligned}$
For second line, $n_1=2$ and $n_2=4$
$$
\begin{aligned}
\frac{1}{\lambda} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{4^2}ight] \\
& =\frac{3}{16} R_{\mathrm{H}}...(ii)
\end{aligned}
$$
Dividing Eqs. (i) and (ii) we get,
$$
\begin{aligned}
\frac{\lambda}{656} & =\frac{5}{36} \times \frac{16}{3} \\
\lambda & =485.9 \mathrm{~nm}
\end{aligned}
$$
Similarly, wavelength of limiting line i.e. $n_2=\infty$ and $n_1=2$
$$
\begin{aligned}
\frac{1}{\lambda} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{\infty^2}ight]=\frac{R_{\mathrm{H}}}{4}=\frac{109737}{4} \\
\therefore \quad \lambda & =364.4 \mathrm{~nm}
\end{aligned}
$$
Hence, wavelength of second line is 485.9 nm and limiting line is 364.4 nm.
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