What is the maximum wavelength line in the Ly | vedclass

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(D) For $He^{+}$ ion,the atomic number $Z = 2$.The Rydberg formula for the wave number is $\bar{
u} = \frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2

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(D) For $He^{+}$ ion,the atomic number $Z = 2$.The Rydberg formula for the wave number is $\bar{
u} = \frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$.For the maximum wavelength $(\lambda_{\max})$,the energy difference must be minimum,which corresponds to the transition from the nearest higher energy level,$n_2 = 2$.Substituting these values: $\frac{1}{\lambda_{\max}} = R (2)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = 4R \left[ 1 - \frac{1}{4} ight] = 4R \left( \frac{3}{4} ight) = 3R$.Therefore,$\lambda_{\max} = \frac{1}{3R}$.

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

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