Lyman vs Balmer Series — Wavelength Ratio Trick in 60 Sec! 🔥

❓ Question

For a hydrogen atom, the ratio of the largest wavelength of the Lyman series to that of the Balmer series is equal to ?

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🖼️ Question Image

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✍️ Short Solution

This is a pure concept + formula-based JEE question.
The key is to remember:

Largest wavelength ⇔ smallest energy difference

So we must identify the closest transition in each series.

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🔹 Step 1 — Rydberg formula (foundation 💯)**

For hydrogen spectrum:

$$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2}),n_2>n_1$$

Where:

• $n_1$ = lower energy level (series identifier)

• $n_2$ = higher energy level

📌 Larger wavelength ⇒ smaller value of $\frac{1}{\lambda}$.

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🔹 Step 2 — Largest wavelength of Lyman series

Lyman series:

$$n_1=1$$

Largest wavelength occurs for smallest jump:

$$n_2=2\to 1$$

So:

$$\frac{1}{{\lambda }_L}=R(1-\frac{1}{4})=\frac{3R}{4}$$

Thus:

$${\lambda }_L=\frac{4}{3R}$$

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🔹 Step 3 — Largest wavelength of Balmer series

Balmer series:

$$n_1=2$$

Largest wavelength ⇒ nearest upper level:

$$n_2=3\to 2$$

So:

$$\frac{1}{{\lambda }_B}=R(\frac{1}{4}-\frac{1}{9})=R\left(\frac{5}{36}\right)$$

Thus:

$${\lambda }_B=\frac{36}{5R}$$

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🔹 Step 4 — Take the ratio

$$\frac{{\lambda }_L}{{\lambda }_B}=\frac{\frac{4}{3R}}{\frac{36}{5R}}$$

Cancel $R$:

$$=\frac{4}{3}\times \frac{5}{36}=\frac{20}{108}=\frac{5}{27}$$

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✅ Final Answer

$$\boxed{{\displaystyle \frac{5}{\mathrm{27}}}}$$

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🔥 Important JEE Insight

• Largest wavelength always corresponds to the smallest energy gap

• For any series:

  • Lyman → $2 \to 1$
    

  • Balmer → $3 \to 2$
    

  • Paschen → $4 \to 3$
    

📌 JEE often tests whether students confuse:

• Largest wavelength ❌ with series limit

• Smallest transition ✔️ (correct logic)