Photon Energy in Atomic Transitions
For a transition between stationary states, conservation of energy gives h nu = hc/lambda = |E_i - E_f|. Emission has E_i > E_f; absorption has E_f > E_i.
Why this shows up in the exam
Finding emitted wavelength · Identifying allowed absorption · Combining atomic photons with photoelectric or collision processes
Learn the idea
An atomic photon carries exactly the magnitude of the energy-level difference. A downward jump releases the missing energy as a photon; an upward jump can absorb a photon only when its energy matches the allowed gap.
🧠 Memory hook: Find the level gap first; only then convert gap to color.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- h nu = h c/lambda = |E_i - E_f| — photon energy for one atomic transition
- nu = c/lambda — vacuum frequency-wavelength relation
How to approach it
- 1Write both level energies
- 2Take the positive magnitude of their difference
- 3Convert consistently among eV, frequency, and wavelength
Common slip-ups that cost marks
- •Using the energy of one level instead of the difference
- •Keeping a negative photon energy
- •Adding wavelengths when energies should add
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.
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