(C) For the $Balmer$ series,$n_1 = 2$ and for the limiting line,$n_2 = \infty$. The Rydberg formula for frequency is $\nu = R_\infty \times Z^2 \times

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$8.22 \times 10^{14} \text{ s}^{-1}$

(C) For the $Balmer$ series,$n_1 = 2$ and for the limiting line,$n_2 = \infty$. The Rydberg formula for frequency is $\nu = R_\infty \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. Substituting $Z = 1$,$n_1 = 2$,and $n_2 = \infty$: $\nu = 3.29 \times 10^{15} \times (\frac{1}{2^2} - \frac{1}{\infty^2}) \text{ s}^{-1}$. $\nu = 3.29 \times 10^{15} \times \frac{1}{4} \text{ s}^{-1} = 8.225 \times 10^{14} \text{ s}^{-1}$.

Class 11 Chemistry Structure of Atom Atomic models and Planck's quantum theory

Medium

For the $H$ spectrum,the limiting line in the $Balmer$ series has a frequency of ...... $(R_\infty = 3.29 \times 10^{15} \text{ cycles/s})$

A
$3.65 \times 10^{14} \text{ s}^{-1}$
B
$3.29 \times 10^{15} \text{ s}^{-1}$
C
$8.22 \times 10^{14} \text{ s}^{-1}$
D
$-8.22 \times 10^{14} \text{ s}^{-1}$

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For the $H$ spectrum,the limiting line in the $Balmer$ series has a frequency of ...... $(R_\infty = 3.29 \times 10^{15} \text{ cycles/s})$

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