What will be the ratio of the wavelength of t | vedclass

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

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$125 : 84$

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(D) The Rydberg formula 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,$n_1 = 2$.For the third line of the Balmer series,$n_2 = 2 + 3 = 5$. For $He^{+}$,$Z = 2$. So,$\frac{1}{\lambda_1} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{5^2} ight) = 4R \left( \frac{1}{4} - \frac{1}{25} ight) = 4R \left( \frac{21}{100} ight) = \frac{21R}{25}$.For the first line of the Balmer series,$n_2 = 2 + 1 = 3$. For $Li^{2+}$,$Z = 3$. So,$\frac{1}{\lambda_2} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 9R \left( \frac{1}{4} - \frac{1}{9} ight) = 9R \left( \frac{5}{36} ight) = \frac{5R}{4}$.The ratio $\frac{\lambda_1}{\lambda_2} = \frac{\lambda_1}{1} 	imes \frac{1}{\lambda_2} = \frac{25}{21R} 	imes \frac{5R}{4} = \frac{125}{84}$.

What will be the ratio of the wavelength of the third line of the Balmer series in the $He^{+}$ ion spectrum to the first line of the Balmer series in the $Li^{2+}$ ion spectrum?

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