An electron in hydrogen transitions from n=3 (E = −1.51 eV) to n=1 (E = −13.6 eV). What is emitted and with what energy?
AA photon absorbed with energy 12.09 eV
BA photon emitted with energy 12.09 eV
CA photon absorbed with energy 15.11 eV
DA photon emitted with energy 1.51 eV
Transitioning from n=3 to n=1 is a downward transition (higher to lower energy), so the electron loses energy by emitting a photon. The photon energy equals the difference: −1.51 − (−13.6) = 12.09 eV. This ultraviolet photon belongs to the Lyman series. Absorption occurs only for upward transitions.
Question 2 True / False
The Bohr model fails for helium because Bohr made an arithmetic error; the underlying physics is correct for most atoms.
TTrue
FFalse
Answer: False
The Bohr model fails for helium because of a fundamental physical limitation, not an arithmetic one. It assumes a single electron experiencing only Coulomb attraction from the nucleus. With two electrons, electron-electron repulsion introduces interactions the model cannot handle. The model's semi-classical orbit picture also cannot predict transition probabilities or fine structure even for hydrogen.
Question 3 Short Answer
What two physical conditions does Bohr combine to derive the allowed orbital radii in hydrogen?
Think about your answer, then reveal below.
Model answer: Bohr combines (1) the balance between Coulomb attraction and centripetal force for a circular orbit, and (2) the postulate that angular momentum is quantized as L = nℏ. Solving these two equations simultaneously yields the discrete allowed radii r_n = n²a₀.
The circular orbit condition (Coulomb = centripetal) alone would allow any radius. The angular momentum quantization condition restricts radius and speed to a discrete set. Only by combining both conditions does Bohr obtain the quantized energy levels E_n = −13.6 eV/n². This two-condition approach is the core derivation strategy.