Q.17 Given: The potential energy of two electrons separated by the Bohr radius is 27.211 eV. The first Bohr radius of hydrogen is 0.5292 Å. The electron makes an orbit of radius 0.5295 Å around the nucleus in hydrogen.
The calculated ionization energy (in eV) of the hydrogen atom is __________.
Introduction
The calculated ionization energy of the hydrogen atom for the given data is
13.6 eV (within rounding). This result follows directly from
Bohr’s model of the hydrogen atom and the physical meaning of the given Coulomb
potential energy data.
Concept and Data Used
For a hydrogen atom in the ground state, the electron occupies the first Bohr orbit
of radius
a0 ≈ 0.529 Å, and its ionization energy is known to be
13.6 eV.
The question provides the following data:
- Potential energy of two electrons separated by the Bohr radius:
U = 27.211 eV
(this corresponds to the Coulomb interaction energy at distance a0,
also known as the Hartree energy). - First Bohr radius of hydrogen:
a0 = 0.5292 Å. - Actual electron orbit radius in this problem:
r = 0.5295 Å,
which is extremely close to a0.
Bohr Model Background
For a hydrogen atom in Bohr’s model, the total energy in the first orbit is:
The ionization energy is the energy required to move the electron from
n = 1 to n = ∞, which equals:
Step-by-Step Reasoning
Step 1: Interpretation of the Given Coulomb Energy
The given electron–electron potential energy at separation a0,
27.211 eV, corresponds to the Coulomb interaction
k e2 / a0.
This value represents the fundamental Coulomb energy scale (Hartree energy).
Step 2: Electron–Proton Interaction in Hydrogen
In a hydrogen atom, the electron experiences an attractive Coulomb potential:
−k e2 / r
At r = a0, the magnitude of this potential energy is the same Coulomb
scale as above, but with a negative sign due to attraction.
Step 3: Energy Relations in Bohr Orbits
In Bohr’s model:
- Kinetic energy K = −E
- Potential energy U = 2E
- Total energy E = K + U
For hydrogen in the ground state (n = 1), this gives:
Step 4: Effect of Slight Radius Difference
The given radius 0.5295 Å differs negligibly from the Bohr radius
0.5292 Å. At the precision of the given data, this difference does
not significantly alter the energy.
Therefore, the ionization energy remains effectively unchanged.
Correct Answer
How Typical Exam Options Are Interpreted
| Option | Value | Explanation |
|---|---|---|
| (A) | 13.6 eV | Correct ground-state ionization energy of hydrogen. |
| (B) | 27.2 eV | Hartree energy; confuses Coulomb scale with binding energy. |
| (C) | 54.4 eV | Applies to hydrogen-like ions such as He+ (Z² × 13.6 eV). |
| (D) | 6.8 eV | Half the correct value due to misapplication of Bohr relations. |
Final Conclusion
Since the given radius is essentially the Bohr radius and the Coulomb energy scale
matches standard hydrogen parameters, the calculated ionization energy of the
hydrogen atom is:
