Let the series limit for the Balmer series be | vedclass

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Question

(B) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.For the Balmer series,$n_{1} = 2$

Vedclass · Accepted Answer

$\lambda_{1} = 0.09 \lambda_{2}$

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(B) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ight)$.For the Balmer series,$n_{1} = 2$. The series limit occurs at $n_{2} = \infty$. Thus,$\frac{1}{\lambda_{1}} = R \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} ight) = \frac{R}{4}$,which implies $\lambda_{1} = \frac{4}{R}$.For the Brackett series,$n_{1} = 4$. The longest wavelength occurs at $n_{2} = 5$. Thus,$\frac{1}{\lambda_{2}} = R \left( \frac{1}{4^{2}} - \frac{1}{5^{2}} ight) = R \left( \frac{1}{16} - \frac{1}{25} ight) = R \left( \frac{25 - 16}{400} ight) = \frac{9R}{400}$.This implies $\lambda_{2} = \frac{400}{9R}$.Now,taking the ratio: $\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} 	imes \frac{R}{4} = \frac{100}{9} = 11.11$ (This seems to be a calculation check,let's re-evaluate).Wait,$\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} / \frac{4}{R} = \frac{400}{36} = 11.11$. Let's re-check the ratio $\frac{\lambda_{1}}{\lambda_{2}} = \frac{4}{R} 	imes \frac{9R}{400} = \frac{36}{400} = 0.09$.Therefore,$\lambda_{1} = 0.09 \lambda_{2}$.

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

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