The shortest wavelength in the Lyman series i | vedclass

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(C) For the $H$-atom, the Rydberg formula for the Lyman series is given by:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)$For t

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$1216$

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(C) For the $H$-atom, the Rydberg formula for the Lyman series is given by:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} ight)$For the shortest wavelength, $n_2 = \infty$:$\frac{1}{\lambda_{	ext{min}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R(1 - 0) = R$Given $\lambda_{	ext{min}} = 912 \ 	ext{Å}$, so $R = \frac{1}{912} \ 	ext{Å}^{-1}$.For the longest wavelength, the transition occurs from the nearest energy level, i.e., $n_2 = 2$:$\frac{1}{\lambda_{	ext{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = R \left( \frac{3}{4} ight)$Substituting the value of $R$:$\frac{1}{\lambda_{	ext{max}}} = \frac{1}{912} 	imes \frac{3}{4} = \frac{3}{3648} = \frac{1}{1216}$Therefore, $\lambda_{	ext{max}} = 1216 \ 	ext{Å}$.

The shortest wavelength in the Lyman series is $912 \ \text{Å}$. Then, the longest wavelength in the series must be (in $\text{Å}$)

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