The lines in the Balmer series have their wav | vedclass

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(C) The wavelength of spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_

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$3647 \,\mathring A$ to $6563 \,\mathring A$

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(C) The wavelength of spectral lines in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R \approx 1.097 	imes 10^7 \, m^{-1}$.For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots$For the longest wavelength (minimum energy),we take $n_2 = 3$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.$\lambda_{\max} = \frac{36}{5R} \approx 6563 \,\mathring A$.For the shortest wavelength (series limit,maximum energy),we take $n_2 = \infty$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty} ight) = \frac{R}{4}$.$\lambda_{\min} = \frac{4}{R} \approx 3647 \,\mathring A$.Thus,the wavelengths lie between $3647 \,\mathring A$ and $6563 \,\mathring A$.

The lines in the Balmer series have their wavelengths lying between

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