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(D) For the Lyman series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$ where $n = 2, 3, 4, \dots$.

Vedclass · Accepted Answer

$9.6$

Vedclass · Answer

(D) For the Lyman series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} ight)$ where $n = 2, 3, 4, \dots$. Shortest wavelength $(\lambda_{L, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{L, min}} = R \implies \lambda_{L, min} = \frac{1}{R}$. Longest wavelength $(\lambda_{L, max})$ occurs at $n = 2$: $\frac{1}{\lambda_{L, max}} = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_{L, max} = \frac{4}{3R}$. $\Delta \lambda_L = \lambda_{L, max} - \lambda_{L, min} = \frac{4}{3R} - \frac{1}{R} = \frac{1}{3R}$. For the Balmer series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$ where $n = 3, 4, 5, \dots$. Shortest wavelength $(\lambda_{B, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{B, min}} = \frac{R}{4} \implies \lambda_{B, min} = \frac{4}{R}$. Longest wavelength $(\lambda_{B, max})$ occurs at $n = 3$: $\frac{1}{\lambda_{B, max}} = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36} \implies \lambda_{B, max} = \frac{36}{5R}$. $\Delta \lambda_B = \lambda_{B, max} - \lambda_{B, min} = \frac{36}{5R} - \frac{4}{R} = \frac{36 - 20}{5R} = \frac{16}{5R}$. Therefore,$\frac{\Delta \lambda_B}{\Delta \lambda_L} = \frac{16/5R}{1/3R} = \frac{16}{5} 	imes 3 = \frac{48}{5} = 9.6$.

If $\Delta \lambda_L$ is the difference between the shortest and longest wavelengths of the Lyman series and $\Delta \lambda_B$ is the difference between the shortest and longest wavelengths of the Balmer series,then $\frac{\Delta \lambda_B}{\Delta \lambda_L} = $

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