The wavelength of the first line of the Balme | vedclass

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

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$1.09 	imes 10^7 \; m^{-1}$

Vedclass · Answer

(A) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first line of the Balmer series, $n_1 = 2$ and $n_2 = 3$.Substituting the values: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36}$.Given $\lambda = 6563 \; Å = 6563 	imes 10^{-10} \; m$.Rearranging for $R$: $R = \frac{36}{5 \lambda} = \frac{36}{5 	imes 6563 	imes 10^{-10}} \approx 1.097 	imes 10^7 \; m^{-1}$.Thus, the Rydberg constant is approximately $1.09 	imes 10^7 \; m^{-1}$.

The wavelength of the first line of the Balmer series is $6563 \; Å$. The Rydberg constant for hydrogen is about

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