In the Balmer series,the wavelength of the 2^ | vedclass

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(C) The Rydberg formula for spectral series is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Balmer se

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$7: 27$

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(C) The Rydberg formula for spectral series is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$. The $2^{	ext{nd}}$ line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.$\frac{1}{\lambda_1} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R Z^2 \left( \frac{1}{4} - \frac{1}{16} ight) = R Z^2 \left( \frac{3}{16} ight)$.Thus,$\lambda_1 = \frac{16}{3 R Z^2}$.For the Paschen series,$n_1 = 3$. The $1^{	ext{st}}$ line corresponds to the transition from $n_2 = 4$ to $n_1 = 3$.$\frac{1}{\lambda_2} = R Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R Z^2 \left( \frac{1}{9} - \frac{1}{16} ight) = R Z^2 \left( \frac{7}{144} ight)$.Thus,$\lambda_2 = \frac{144}{7 R Z^2}$.Taking the ratio $\frac{\lambda_1}{\lambda_2} = \left( \frac{16}{3 R Z^2} ight) 	imes \left( \frac{7 R Z^2}{144} ight) = \frac{16 	imes 7}{3 	imes 144} = \frac{112}{432} = \frac{7}{27}$.

In the Balmer series,the wavelength of the $2^{\text{nd}}$ line is $\lambda_1$ and for the Paschen series,the wavelength of the $1^{\text{st}}$ line is $\lambda_2$. Then the ratio $\lambda_1 : \lambda_2$ is:

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