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8 chem 107 class

# 8 chem 107 class - PHOTOELECTRIC EFFECT Light in Electrons...

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1 PHOTOELECTRIC EFFECT Light in Electrons out Metal Surface DETECT NUMBER OF ELECTRONS AND THEIR KINETIC ENERGY OBSERVATIONS # e # e ν I KE ν KE I ν 0 ν 0 ν > ν 0 ν > ν 0 PHOTON – a “particle” of light ENERGY OF A PHOTON E = h ν where h is the Planck’s constant: h = 6.626 x 10 -34 J s LIGHT AS A PARTICLE LIGHT AS A PARTICLE ENERGY AND WAVELENGTH E = h ν and c = λ ν , then E = (h c) / λ APPLICATION SAMPLE CALCULATION A laser emits red light with a wavelength of 633 nm, what is the energy of the photons being released at this wavelength? APPLICATION SAMPLE CALCULATION Given: wavelength = 633 nm We know: E = h ν and c = λ ν so: E = (h c) / λ

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2 APPLICATION SAMPLE CALCULATION E = (h c) / λ = (6.626 x 10 -34 J s) x (3 x 10 8 m s -1 ) 633 x 10 -9 m = 3.14 x 10 -19 J SAMPLE CALCULATION – 2 The particular laser describe before has an output of 1 mW (1 x 10 -3 J s -1 ). How many photons are emitted per second? no. of photons = Output (power) / E = (1 x 10 -3 J s -1 ) (3.14 x 10- 19 J / photon) = 3.18 x 10 15 photons / s APPLICATION ENERGY OF A PHOTON E = h ν LIGHT AS A PARTICLE # e ν ν 0 THRESHOLD FREQUENCY Energy greater than E ν 0 is required to eject an electron E ν 0 = h ν 0 BINDING ENERGY the force that atoms exert to hold onto their electrons So, part of the energy of the photon is absorbed to overcome the binding energy. As ν is increased and ν > ν 0 part of the energy is still absorbed to overcome the binding energy
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