Hydrogen-like atom – ground-state binding energy In its lowest (ground) state, the electron bound to a proton (hydrogen atom) has a binding/ionization energy equal to which of the following?
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A100.5 eV
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B51.2 eV
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C13.6 eV
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D0.2 eV
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E1.6 eV
Answer
Correct Answer: 13.6 eV
Explanation
Introduction:The Bohr model and quantum mechanics both predict discrete energy levels for the hydrogen atom. The ground-state energy determines the photon energy required to ionize hydrogen from n = 1 to the continuum, a benchmark value in spectroscopy and astrophysics.
Given Data / Assumptions:
- Hydrogen atom (one electron, one proton) at rest.
- Energy levels depend on principal quantum number n.
- Binding energy is the energy needed to remove the electron to infinity (zero reference).
Concept / Approach:
In the Bohr model, allowed energies are E_n = −13.6 eV / n^2. For the ground state n = 1, E_1 = −13.6 eV. The ionization energy is the magnitude required to raise the electron to zero energy, hence 13.6 eV. This value matches precise spectroscopic measurements and underpins the Rydberg constant.
Step-by-Step Solution:
Use E_n = −13.6 eV / n^2 for hydrogen-like atoms.Set n = 1 → E_1 = −13.6 eV.Ionization (binding) energy magnitude is |E_1| = 13.6 eV.Verification / Alternative check:
The Lyman series limit corresponds to 13.6 eV photons (≈ 91.2 nm), consistent with this binding energy.
Why Other Options Are Wrong:
51.2 eV equals 4 × 12.8 eV, not a hydrogen ground-state value; 100.5 eV and 1.6 or 0.2 eV are inconsistent with the known spectrum.
Common Pitfalls:
Confusing excitation energies (e.g., Balmer lines) with ionization energy; forgetting sign convention for bound-state energies.
Final Answer:
13.6 eV