In a hydrogen spectrum,let lambda be the wave | vedclass

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(A) For the first line of the Lyman series $(n_1=1, n_2=2)$:$\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4

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$5$

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(A) For the first line of the Lyman series $(n_1=1, n_2=2)$:$\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}ight) = R \left(1 - \frac{1}{4}ight) = \frac{3R}{4} \implies \lambda = \frac{4}{3R} \quad \dots(1)$For the $3^{	ext{rd}}$ line of the Paschen series $(n_1=3, n_2=6)$:$\frac{1}{\lambda_3} = R \left(\frac{1}{3^2} - \frac{1}{6^2}ight) = R \left(\frac{1}{9} - \frac{1}{36}ight) = R \left(\frac{4-1}{36}ight) = \frac{3R}{36} = \frac{R}{12} \implies \lambda_3 = \frac{12}{R} \quad \dots(2)$For the $2^{	ext{nd}}$ line of the Balmer series $(n_1=2, n_2=4)$:$\frac{1}{\lambda_2} = R \left(\frac{1}{2^2} - \frac{1}{4^2}ight) = R \left(\frac{1}{4} - \frac{1}{16}ight) = R \left(\frac{4-1}{16}ight) = \frac{3R}{16} \implies \lambda_2 = \frac{16}{3R} \quad \dots(3)$The wavelength difference is given by $\lambda_3 - \lambda_2 = a\lambda$:$a\lambda = \frac{12}{R} - \frac{16}{3R} = \frac{36 - 16}{3R} = \frac{20}{3R}$Substituting $\lambda = \frac{4}{3R}$ into the equation:$a \left(\frac{4}{3R}ight) = \frac{20}{3R} \implies a = \frac{20}{4} = 5$

In a hydrogen spectrum,let $\lambda$ be the wavelength of the first transition line of the Lyman series. The wavelength difference between the $3^{\text{rd}}$ transition line of the Paschen series and the $2^{\text{nd}}$ transition line of the Balmer series is $a\lambda$,where $a = ........$

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