The second line of the Balmer series has a wa | vedclass

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(B) The wavelength of the spectral lines in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{

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$6563$

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(B) The wavelength of the spectral lines in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$,where $n = 3, 4, 5, \dots$For the second line of the Balmer series,$n = 4$. Given $\lambda_2 = 4861 Å$:$\frac{1}{4861} = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{3}{16} ight) \implies R = \frac{16}{3 	imes 4861} \dots (i)$For the first line of the Balmer series,$n = 3$:$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$Substituting the value of $R$ from equation $(i)$:$\frac{1}{\lambda_1} = \left( \frac{16}{3 	imes 4861} ight) 	imes \left( \frac{5}{36} ight) = \frac{80}{108 	imes 4861} = \frac{20}{27 	imes 4861}$$\lambda_1 = \frac{27 	imes 4861}{20} = \frac{131247}{20} = 6562.35 Å \approx 6563 Å$.

The second line of the Balmer series has a wavelength of $4861 Å$. The wavelength of the first line of the Balmer series is: (in $Å$)

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