The first three spectral lines of the H-atom | vedclass

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(D) For the $1^{	ext{st}}$ line of the Balmer series $(n_1 = 2, n_2 = 3)$:$\frac{1}{\lambda_{1}} = R \left(\frac{1}{2^{2}} - \frac{1}{3^{2}}ight) =

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$15$

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(D) For the $1^{	ext{st}}$ line of the Balmer series $(n_1 = 2, n_2 = 3)$:$\frac{1}{\lambda_{1}} = R \left(\frac{1}{2^{2}} - \frac{1}{3^{2}}ight) = R \left(\frac{1}{4} - \frac{1}{9}ight) = R \left(\frac{5}{36}ight) \implies \lambda_{1} = \frac{36}{5R}$.For the $3^{	ext{rd}}$ line of the Balmer series $(n_1 = 2, n_2 = 5)$:$\frac{1}{\lambda_{3}} = R \left(\frac{1}{2^{2}} - \frac{1}{5^{2}}ight) = R \left(\frac{1}{4} - \frac{1}{25}ight) = R \left(\frac{21}{100}ight) \implies \lambda_{3} = \frac{100}{21R}$.Calculating the ratio $\frac{\lambda_{1}}{\lambda_{3}}$:$\frac{\lambda_{1}}{\lambda_{3}} = \left(\frac{36}{5R}ight) 	imes \left(\frac{21R}{100}ight) = \frac{36 	imes 21}{500} = \frac{756}{500} = 1.512$.Expressing as $x 	imes 10^{-1}$:$1.512 = 15.12 	imes 10^{-1}$.Rounding to the nearest integer,$x \approx 15$.

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