Taking the wavelength of the first Balmer lin | vedclass

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(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For the first B

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$488.9$

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(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first Balmer line $(n_i = 3, n_f = 2)$: $\frac{1}{660} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$ .... $(1)$For the second Balmer line $(n_i = 4, n_f = 2)$: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = \frac{3R}{16}$ .... $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\lambda}{660} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$$\lambda = 660 	imes \frac{20}{27} = \frac{13200}{27} \approx 488.88\,nm \approx 488.9\,nm$.

Taking the wavelength of the first Balmer line in the hydrogen spectrum ($n = 3$ to $n = 2$) as $660\,nm$,the wavelength of the $2^{nd}$ Balmer line ($n = 4$ to $n = 2$) will be....$nm$.

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