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Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
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1904 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{27}$
Wavelength for Lyman series
$$
\lambda_L=\frac{1}{R\left(1-\frac{1}{4}\right)}=\frac{4}{3 R}
$$
and wavelength for Balmer series
$$
\begin{aligned}
& \lambda_B=\frac{1}{R\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{1}{R\left(\frac{5}{36}\right)}=\frac{36}{5 R} \\
& \therefore \quad \frac{\lambda_L}{\lambda_B}=\frac{4}{3 R} \times \frac{5 R}{36}=\frac{5}{27} \\
& \Rightarrow \quad \lambda_L: \lambda_B=5: 27 \\
&
\end{aligned}
$$
$$
\lambda_L=\frac{1}{R\left(1-\frac{1}{4}\right)}=\frac{4}{3 R}
$$
and wavelength for Balmer series
$$
\begin{aligned}
& \lambda_B=\frac{1}{R\left(\frac{1}{4}-\frac{1}{9}\right)}=\frac{1}{R\left(\frac{5}{36}\right)}=\frac{36}{5 R} \\
& \therefore \quad \frac{\lambda_L}{\lambda_B}=\frac{4}{3 R} \times \frac{5 R}{36}=\frac{5}{27} \\
& \Rightarrow \quad \lambda_L: \lambda_B=5: 27 \\
&
\end{aligned}
$$
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