2nd Excited State → Find Atomic Number FAST ⚡

 

❓ Question

In a hydrogen-like ion, the energy difference between the 2nd excited state and the ground state is 108.8 eV.

Find the atomic number (Z) of the ion.

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🖼️ Question Image

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✍️ Short Solution

This is a direct Bohr-model energy-level question.
No tricks, no approximations — just n-value clarity + Z² scaling 🔥

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🔹 Step 1 — Understand ‘2nd Excited State’ (MOST IMPORTANT 💯)**

Energy levels in hydrogen-like ions:

• Ground state → $n = 1$
  

• 1st excited state → $n = 2$
  

• 2nd excited state → $n = 3$ ✅

📌 Many students mess up here — 2nd excited ≠ n = 2

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🔹 Step 2 — Energy Formula for Hydrogen-like Ion

Bohr energy level:

$$E_n=-13.6\frac{Z^2}{n^2}\text{ eV}$$

Where:

• $Z$ = atomic number

• $n$ = principal quantum number

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🔹 Step 3 — Write Energies of Required Levels

Ground state ($n=1$):

$$E_1=-13.6Z^2$$

2nd excited state ($n=3$):

$$E_3=-\frac{13.6Z^2}{9}$$

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🔹 Step 4 — Energy Difference Given

Energy difference:

$$\mathrm{\Delta }E=E_3-E_1$$

Taking magnitude:

$$\Delta E = 13.6 Z^2\left(1 - \frac{1}{9}\right)$$
$$\Delta E = 13.6 Z^2 \cdot \frac{8}{9}$$

Given:

$$\Delta E = 108.8\ \text{eV}$$

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🔹 Step 5 — Solve for Z

$$13.6 \times \frac{8}{9} \times Z^2 = 108.8$$

Multiply both sides by 9:

$$13.6 \times 8 \times Z^2 = 979.2$$
$$108.8 \times Z^2 = 979.2$$
$$Z^2 = 9$$
$$\boxed{Z = 3}$$

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✅ Final Answer

$$\boxed{Z = 3}$$

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⭐ Golden JEE Insight

• Hydrogen-like ion ⇒ Z² scaling

• 2nd excited state ⇒ n = 3 (always!)

• Energy difference formula:

$$\Delta E = 13.6 Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

🧠 One-line memory:

Excitation number + 1 = principal quantum number

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