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

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$4860$

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(C) The wavelength of a spectral line in the Balmer series is given by the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} ight]$For the first line of the Balmer series,$n = 3$:$\frac{1}{\lambda_1} = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{5}{36} ight]$Given $\lambda_1 = 6561 \; \mathring{A}$,so $\frac{1}{6561} = \frac{5R}{36} \implies R = \frac{36}{5 	imes 6561}$.For the second line of the Balmer series,$n = 4$:$\frac{1}{\lambda_2} = R \left[ \frac{1}{4} - \frac{1}{16} ight] = R \left[ \frac{3}{16} ight]$Substituting the value of $R$:$\frac{1}{\lambda_2} = \left( \frac{36}{5 	imes 6561} ight) 	imes \frac{3}{16} = \frac{108}{80 	imes 6561} = \frac{27}{20 	imes 6561}$$\lambda_2 = \frac{20 	imes 6561}{27} = 20 	imes 243 = 4860 \; \mathring{A}$.

If the wavelength of the first line of the Balmer series of hydrogen is $6561 \; \mathring{A}$,the wavelength of the second line of the series should be........$\mathring{A}$

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