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10 cards from this deck
Light on metal surface ejects electrons
Min. frequency needed to emit photoelectrons from metal
Energy electron gains/loses moving through 1V potential diff.
1.6×10−191.6 \times 10^{-19}1.6×10−19 J
E=hfE = hfE=hf or E=hcλE = \frac{hc}{\lambda}E=λhc
6.63×10−346.63 \times 10^{-34}6.63×10−34 J s
Ek max=hf−ϕE_{k \text{ max}} = hf - \phiEk max=hf−ϕ
Min. energy required to release photoelectron from metal
More photoelectrons emitted, but same max. kinetic energy
Min. voltage to just stop most energetic photoelectrons
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