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The limiting line in Balmer series will have a frequency of (Rydberg constant, $R_{\infty}=3.29 \times 10^{15}$ cycles $/ \mathrm{s}$ )
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Verified Answer
The correct answer is:
$8.22 \times 10^{14} \mathrm{~s}^{-1}$
$8.22 \times 10^{14} \mathrm{~s}^{-1}$
$\bar{v}=\frac{1}{\lambda}=R_H Z\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
In Balmer series $n_1=2 \& n_2=3,4,5 \ldots$... Last line of the spectrum is called series limit.
Limiting line is the line of shortest wavelength and high energy when $n_2=\infty$
$$
\begin{aligned}
\therefore \quad \bar{v}= & \frac{1}{\lambda}=\frac{R_H}{n_1^2}=\frac{3.29 \times 10^{15}}{2^2}=\frac{3.29 \times 10^{15}}{4} \\
& =8.22 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$
In Balmer series $n_1=2 \& n_2=3,4,5 \ldots$... Last line of the spectrum is called series limit.
Limiting line is the line of shortest wavelength and high energy when $n_2=\infty$
$$
\begin{aligned}
\therefore \quad \bar{v}= & \frac{1}{\lambda}=\frac{R_H}{n_1^2}=\frac{3.29 \times 10^{15}}{2^2}=\frac{3.29 \times 10^{15}}{4} \\
& =8.22 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$
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