Let lambda_P and lambda_L be the longest wave | vedclass

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(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \r

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$15 < \frac{\lambda_P}{\lambda_L} < 16$

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(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Paschen series,$n_1 = 3$. The longest wavelength corresponds to the transition from $n_2 = 4$.$\frac{1}{\lambda_P} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{9} - \frac{1}{16} ight] = R \left[ \frac{16 - 9}{144} ight] = \frac{7R}{144}$.Thus,$\lambda_P = \frac{144}{7R}$.For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the transition from $n_2 = 2$.$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4}$.Thus,$\lambda_L = \frac{4}{3R}$.Now,calculating the ratio $\frac{\lambda_P}{\lambda_L}$:$\frac{\lambda_P}{\lambda_L} = \frac{144 / 7R}{4 / 3R} = \frac{144}{7R} 	imes \frac{3R}{4} = \frac{36 	imes 3}{7} = \frac{108}{7} \approx 15.42$.Since $15 < 15.42 < 16$,the correct option is $C$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

Let $\lambda_P$ and $\lambda_L$ be the longest wavelengths observed in the Paschen and Lyman series respectively. Choose the correct option.

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