Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n = 4 to n…

🧠 The Energy Matching Game Two transitions Have the same wavelength if they have the same energy gap ($\Delta E$). Since $\Delta E \propto Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, we are looking for a pair where the "effective levels" ($n/Z$) match perfectly.

🗺️ The Algebraic Equalizer

1. The $\ce{He^+}$ Transition: For $\ce{He^+}$, $Z=2$. The transition is $4 \to 2$. $\Delta E_{\ce{He^+}} \propto R(2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right)$
2. Distributing the $Z^2$: $\Delta E_{\ce{He^+}} \propto R \left( \frac{4}{4} - \frac{4}{16} \right) = R \left( \frac{1}{1} - \frac{1}{4} \right)$
3. The Hydrogen Equivalent: This matches the formula for Hydrogen ($Z=1$) with levels $2 \to 1$: $\Delta E_{\text{H}} \propto R(1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right)$ Thus, the $4 \to 2$ line in Helium is exactly the same energy as the $2 \to 1$ (Lyman-alpha) line in Hydrogen.

⚡ The $n/Z$ Shortcut For ions, the energy levels effectively feel like $n_{\text{eff}} = n/Z$. Helium $n=4 \implies$ Hydrogen $n=4/2=2$. Helium $n=2 \implies$ Hydrogen $n=2/2=1$. So, $\ce{He^+} (4 \to 2)$ is identical to $\text{H} (2 \to 1)$. Fast and foolproof!

⚠️ Common Traps Don't mix up the direction! Even though the energy gap is the same, transitions in options are often given as $n_i \to n_f$ or $n_f \to n_i$. Always ensure you are comparing like-for-like (emission vs emission).

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