The shortest wavelength of H atom in the Lyman series is 1. The longest wavelength in the Balmer series of He+ is:

🧠 The Scaling Identity This is a conceptual twin to ATOM-062. We are comparing the "series limit" of Hydrogen's ground state against the "first line" of Helium's second series. Because $Z=2$ for Helium, it exactly cancels out the energy drop of moving to $n=2$, keeping the wavelengths in the same order of magnitude.

🗺️ The Derivation Map

1. Lyman Shortest (H): Transition $\infty \to 1$. $\frac{1}{\lambda_1} = R(1)^2 \left( \frac{1}{1^2} - 0 \right) \implies R = \frac{1}{\lambda_1}$
2. Balmer Longest ($\ce{He^+}$): Transition $3 \to 2$ with $Z=2$. $\frac{1}{\lambda_{max}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right)$
3. Solving: $\frac{1}{\lambda_{max}} = \frac{5R}{9} = \frac{5}{9\lambda_1} \implies \lambda_{max} = \frac{9\lambda_1}{5}$

⚡ The 9:5 Ratio If you memorize that the $H_\alpha$ line factor is $5/36$, and the Rydberg limit factor is $1$, the rest is just algebra with $Z^2$. For Helium, the $Z^2$ is $4$. $4 \times (5/36) = 5/9$. Reciprocal factor $= 9/5$.

⚠️ Common Traps Be careful with the variable name! The question uses $\lambda_1$. Don't let your eyes skip the subscript and pick a generic constant.

$\boxed{\text{Answer: (C)}}$