The shortest wavelength of hydrogen atom in Lyman series is . The longest wavelength in Balmer series of He+ is:

🧠 The Inverse Energy Map Wavelength is inversely proportional to energy ($\lambda \propto 1/E$).

• "Shortest wavelength" = Maximum Energy ($n = \infty \to n_f$).
• "Longest wavelength" = Minimum Energy (the very first jump, $n_i = n_f + 1 \to n_f$). We scale Hydrogen's Lyman limit to find $\ce{He^+}$'s Balmer start.

🗺️ The Mathematical Bridge

1. Hydrogen Lyman Limit ($\lambda$): The transition is $n = \infty \to 1$ with $Z=1$. $\frac{1}{\lambda} = R(1)^2 \left( \frac{1}{1^2} - 0 \right) = R \implies R = \frac{1}{\lambda}$
2. $\ce{He^+}$ Balmer Longest ($\lambda'$): The transition is $3 \to 2$ with $Z=2$. $\frac{1}{\lambda'} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right)$
3. The Substitution: $\frac{1}{\lambda'} = 4 \left( \frac{1}{\lambda} \right) \left( \frac{5}{36} \right) = \frac{20}{36\lambda} = \frac{5}{9\lambda}$ $\lambda' = \frac{9\lambda}{5}$

⚡ The 30-Second Scaling Rule $\lambda \propto \frac{1}{Z^2 \cdot \text{factor}}$. For H-Lyman Limit: $Z^2 \times \text{factor} = 1^2 \times 1 = 1$. For $\ce{He^+}$-Balmer Longest: $Z^2 \times \text{factor} = 2^2 \times (1/4 - 1/9) = 4 \times 5/36 = 5/9$. Since the factor dropped from $1$ to $5/9$, the wavelength must increase by the reciprocal ($9/5$).

⚠️ Common Traps Watch out for the $Z^2$ term! Many students correctly identify the $3 \to 2$ transition but forget that $\ce{He^+}$ has two protons, which makes the energy gaps $4$ times larger, pulling the wavelength back down.

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