If {lambda _{{text{max}}}} is 6563 text{ Å}, | vedclass

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(A) For the Balmer series, the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$, where $n = 3, 4, 5, \

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$\lambda = \frac{16}{3R}$

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(A) For the Balmer series, the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$, where $n = 3, 4, 5, \dots$ For the first line (maximum wavelength), $n = 3$. For the second line, $n = 4$. Substituting $n = 4$ into the formula: $\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{16} \right)$ $\frac{1}{\lambda} = R \left( \frac{4 - 1}{16} \right) = \frac{3R}{16}$ Therefore, $\lambda = \frac{16}{3R}$.

If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

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