The first member of the Balmer series of a hy | vedclass

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(C) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R_{H} \left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right)$,where $n = 3, 4

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$486$

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(C) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R_{H} \left( \frac{1}{2^{2}} - \frac{1}{n^{2}} ight)$,where $n = 3, 4, 5, \dots$The first member corresponds to $n = 3$: $\frac{1}{\lambda_{1}} = R_{H} \left( \frac{1}{4} - \frac{1}{9} ight) = R_{H} \left( \frac{5}{36} ight)$.The second member corresponds to $n = 4$: $\frac{1}{\lambda_{2}} = R_{H} \left( \frac{1}{4} - \frac{1}{16} ight) = R_{H} \left( \frac{3}{16} ight)$.Taking the ratio: $\frac{\lambda_{2}}{\lambda_{1}} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$.Given $\lambda_{1} = 6561 \; \mathring{A}$,we have $\lambda_{2} = \frac{20}{27} 	imes 6561 = 4860 \; \mathring{A}$.Converting to nanometers: $4860 \; \mathring{A} = 486 \; nm$.

The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is

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