Physics Q44 – JEE-MAIN 2025

In a hydrogen like ion, the energy difference between the

2^{n d}

excitation energy state and ground state is

108.8

eV. The atomic number of the ion is:

Clue 1 of 3: Energy Levels in Hydrogen-like Ions

Recall the formula for the energy of an electron in a hydrogen-like ion and correctly identify the principal quantum numbers for the ground state and the 2nd excitation state.

🥷

Ninja Strategy•Order of Magnitude Check

Estimate the energy difference for each given $Z$ value to quickly eliminate options that yield values too far from the given $108.8$ eV.

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Step 1: Identify Energy Levels✦ Active

The ground state corresponds to the principal quantum number $n_{1} = 1$ . The 2nd excitation energy state means the electron is in the $n = 3$ level (the 1st excited state is $n = 2$ , and the 2nd excited state is $n = 3$ ), so $n_{2} = 3$ .

Step 2: Apply Bohr's Energy Formula○ Expand

The energy of an electron in a hydrogen-like ion with atomic number $Z$ is given by $E_{n} = - 13.6 \frac{Z^{2}}{n^{2}}$ eV. The energy difference between the 2nd excitation state ( $n = 3$ ) and the ground state ( $n = 1$ ) is $Δ E = E_{3} - E_{1}$ .

Δ E = (- 13.6 \frac{Z^{2}}{3^{2}}) - (- 13.6 \frac{Z^{2}}{1^{2}})

💡 Teacher's Secret Hint

Ensure correct signs when calculating the energy difference.

Step 3: Calculate Atomic Number (Z)○ Expand

Substitute the given energy difference $Δ E = 108.8$ eV into the equation and solve for $Z$ :

108.8 = 13.6 Z^{2} (1 - \frac{1}{9})

108.8 = 13.6 Z^{2} (\frac{8}{9})

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

Z^{2} = \frac{108.8 \times 9}{108.8}

Z^{2} = 9

Z = 3

Thus, the atomic number of the ion is $3$ .

💡 Teacher's Secret Hint

Simplify the numerical calculation carefully to avoid errors.

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Physics Question 44 – JEE-MAIN 2025

Clue 1 of 3: Energy Levels in Hydrogen-like Ions

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