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PH 122 (46)

Heisenberg uncertainty principle

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PH 122
Brian Jones

2 April Light of wavelength 400 nm illuminates a potassium electrode (work function 2.3 eV) What is the photon energy? 1240 E photon λ = 3.1 eV What is the energy of the emitted electron? 3.1 eV – 2.3 eV = 0.8 eV What is the stopping potential? -0.8 V Ocean water is most transparent at wavelengths of 470 nm, so bioluminescent creatures emit light at approximately this wavelength. Firefly squid use ATP to provide the energy for this reaction. Metabolizing one molecule of ATP releases 0.32 eV. How many molecules of ATP must be metabolized to produce one photon of blue light at 470 nm? 1240 1240 E photon = = 2.6 eV λ 470nm 2.6eV 0.32eV = 8.15 9 molecules of ATP Electrons are accelerated from rest through an 8000 V potential difference. By what factor would their de Broglie wavelength increase if they were instead accelerated through a 2000 V potential? K = Δ Ue K = ½ mv 2 h h λ = p = mv ΔU ↓ 4 K ↓ 4 v ↓ 2
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