What is the maximum wavelength line in the Ly | vedclass

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

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$\frac{1}{3R}$

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(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} ight)$.For the Lyman series,$n_1 = 1$. For maximum wavelength,the transition must be from the nearest energy level,so $n_2 = 2$.For $He^{+}$ ion,the atomic number $Z = 2$.Substituting these values: $\frac{1}{\lambda_{\max}} = R (2)^{2} \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} ight)$.$\frac{1}{\lambda_{\max}} = R 	imes 4 	imes \left( 1 - \frac{1}{4} ight) = R 	imes 4 	imes \frac{3}{4} = 3R$.Therefore,$\lambda_{\max} = \frac{1}{3R}$.

What is the maximum wavelength line in the Lyman series of $He^{+}$ ion?

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