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QPages 382-383: Q2: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 382-383: Q2 Solution: The first excited energy of a He ion is the same as the ground…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often assume that because $Z$ is an integer, the condition $n=Z$ is always physically valid, but they fail to verify that the principal quantum number $n$ must be a positive integer. While $n=Z$ works for all integer $Z$, the trap lies in recognizing that this specific state exists only if $Z$ is an integer, which is true for all real hydrogen-like ions, but the question's phrasing can mislead students into overgeneralizing without checking the domain of $n$.

The first excited energy of a He+^+ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?

Step-by-Step Verified Solution

**Answer:** No, it is not always true. **Step-by-Step Explanation:** For He_ (Z = 2), the energy of the first excited state (n = 2) is:
E=−13.6×Z2n2=−13.6×44=−13.6 eVE = -13.6 \times \frac{Z^2}{n^2} = -13.6 \times \frac{4}{4} = -13.6 \text{ eV}
which is the same as the ground state of hydrogen. However, for a general hydrogen-like ion with atomic number Z, the energy is En=−13.6Z2/n2E_n = -13.6 Z^2/n^2. For this to equal _13.6 eV, we need Z2/n2=1Z^2/n^2 = 1, i.e., n = Z. This is only possible if Z is an integer, which is true for all hydrogen-like ions. But the question asks if it is *always* true that *one of the energies* will be the same. For any Z, the state with n = Z will have energy _13.6 eV. So yes, it is always true [citation:3].

All Questions in this Chapter

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