Bohr's Model Of Atom
The ionisation energy of a hydrogen-like ion He⁺ in its ground state is (ionisation energy of H = 13.6 eV):
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Solution
54.4 eV
For hydrogen-like species, the energy of an electron in the nth orbit is Eₙ = −13.6 × Z²/n² eV, where Z is the atomic number. For He⁺ (Z = 2) in the ground state (n = 1): E₁ = −13.6 × 4/1 = −54.4 eV. Ionisation energy is the energy required to remove the electron completely (to n = ∞ where E = 0), so IE = 0 − (−54.4) = 54.4 eV. This is 4 times the ionisation energy of hydrogen because IE scales as Z², and for helium Z = 2, so IE(He⁺) = 4 × 13.6 = 54.4 eV.
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About This Question
- Subject
- chemistry
- Chapter
- atomic structure
- Topic
- bohr's model of atom
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
54.4 eV
For hydrogen-like species, the energy of an electron in the nth orbit is Eₙ = −13.6 × Z²/n² eV, where Z is the atomic number. For He⁺ (Z = 2) in the ground state (n = 1): E₁ = −13.6 × 4/1 = −54.4 eV. Ionisation energy is the energy required to remove the electron completely (to n = ∞ where E = 0), so IE = 0 − (−54.4) = 54.4 eV. This is 4 times the ionisation energy of hydrogen because IE scales as Z², and for helium Z = 2, so IE(He⁺) = 4 × 13.6 = 54.4 eV.
This easy difficulty chemistry question is from the chapter atomic structure, covering the topic of bohr's model of atom. It appeared in the 2025 exam.
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