Concept: Wavelength is inversely proportional to energy. Therefore, the spectral series involving transitions in the highest energy levels (smallest energy gaps) produces photons with the lowest energy and largest wavelength.
Formula: $$ \Delta E = \frac{hc}{\lambda} $$
Solution: - The Hydrogen emission spectrum series fall to different base levels:
- Lyman: falls to \( n=1 \) (UV region, huge energy, shortest wavelength)
- Balmer: falls to \( n=2 \) (Visible)
- Paschen: falls to \( n=3 \) (IR)
- Brackett: falls to \( n=4 \) (IR)
- Pfund: falls to \( n=5 \) (Far IR, smallest energy transitions, largest wavelengths).
- Because the energy gap between higher shells (e.g., \( n=6 \) to \( n=5 \)) is extremely small, the emitted photon has minimal energy and maximum wavelength.
Why other options are incorrect: The other series all drop to lower principal quantum numbers, meaning their electron transitions cover larger energy gaps, yielding shorter wavelengths.
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