IAC Competition Prep · 2025

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Physics · Quantum ⚡ Olympiad

An electron in a hydrogen atom transitions from n = 4 to n = 2. Using En = −13.6 / n² eV, what is the energy of the emitted photon?

A1.89 eV
B2.55 eV
C0.85 eV
D3.40 eV
💡 ΔE = E₄ − E₂ = −13.6/16 − (−13.6/4) = −0.85 + 3.40 = 2.55 eV. This photon corresponds to the Hα Balmer series line (486 nm, blue-green).
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