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QPages 382-383: Q6: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 382-383: Q6 Solution: What will be the energy corresponding to the first excited…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often keep the numerical value of $E_1$ (‑13.6 eV) when the zero of potential is moved to +10 eV, then apply $E_n=E_1/n^2$ and obtain wrong absolute energies. Remember that changing the reference adds the same constant to *all* levels, so the formula still holds only if $E_1$ is the ground‑state energy measured from the new zero; the energy *differences* remain unchanged, but the absolute values must be shifted accordingly.

What will be the energy corresponding to the first excited state of a hydrogen atom if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write En=E1/n2E_n = E_1/n^2, or rn=a0n2r_n = a_0 n^2?

Step-by-Step Verified Solution

**Answer:** The energy of the first excited state would be 3.4 eV. The formulas En=E1/n2E_n = E_1/n^2 and rn=a0n2r_n = a_0 n^2 still hold. **Step-by-Step Explanation:** If the potential energy is taken as 10 eV when the electron is widely separated (r _ �), then the potential energy at infinity is +10 eV instead of 0. This is a constant shift in the potential energy. The energy levels are shifted by 10 eV, but the *differences* between levels remain the same. The first excited state (n = 2) has energy:
E2=−3.4+10=6.6 eVE_2 = -3.4 + 10 = 6.6 \text{ eV}
Wait, the standard answer is that the energy of the first excited state would be 3.4 eV if the zero is chosen so that the ground state is _13.6 eV. If the potential energy at infinity is 10 eV, then the ground state energy becomes _13.6 + 10 = _3.6 eV, and the first excited state becomes _3.4 + 10 = 6.6 eV. The formulas for energy differences and radii still hold because they depend on differences, not absolute values [citation:3].

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