What is the maximum wavelength in the Lyman s | vedclass

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(B) For the Lyman series,the transition occurs from higher energy levels to $n_1 = 1$. For maximum wavelength,the energy difference must be minimum,wh

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$\frac{1}{3R}$

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(B) For the Lyman series,the transition occurs from higher energy levels to $n_1 = 1$. For maximum wavelength,the energy difference must be minimum,which corresponds to the transition from $n_2 = 2$ to $n_1 = 1$. The Rydberg formula is $\frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$. For $He^+$ ion,$Z = 2$,so $Z^2 = 4$. Substituting the values: $\frac{1}{\lambda} = R 	imes 4 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 4R \left( 1 - \frac{1}{4} ight) = 4R \left( \frac{3}{4} ight) = 3R$. Therefore,$\lambda = \frac{1}{3R}$.

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

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