The shortest wavelength for the Lyman series | vedclass

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Question

(C) The shortest wavelength in the Lyman series occurs for the transition from $n = \infty$ to $n = 1$:$\frac{1}{\lambda_{L}} = R \left[ \frac{1}{1^2}

Vedclass · Accepted Answer

$18760$

Vedclass · Answer

(C) The shortest wavelength in the Lyman series occurs for the transition from $n = \infty$ to $n = 1$:$\frac{1}{\lambda_{L}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R$$	herefore \lambda_{L} = \frac{1}{R} = 912 \ 	ext{Å}$The longest wavelength in the Paschen series occurs for the transition from $n = 4$ to $n = 3$:$\frac{1}{\lambda_{P}} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{9} - \frac{1}{16} ight] = R \left[ \frac{16 - 9}{144} ight] = \frac{7R}{144}$$	herefore \lambda_{P} = \frac{144}{7R}$Substituting $\frac{1}{R} = 912 \ 	ext{Å}$:$\lambda_{P} = \frac{144}{7} 	imes 912 \ 	ext{Å} \approx 18760 \ 	ext{Å}$

The shortest wavelength for the Lyman series is $912 \ \text{Å}$. The longest wavelength in the Paschen series is: (in $\text{Å}$)

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