Question

9. (15 points) Electrons are emitted from the surface of a metal when it's exposed to light. This is called the photoclectric effect. Each metal has a certain threshold frequency of light, below which nothing happens. Right at this threshold frequency, an electron is emitted. The binding energy of a metal is the minimum amount of energy required to remove an electron from the surface of the metal. The binding energy for lithium metal is 279.7 kJ/mol (that is, it takes 279.7 kJ of energy to remove one mole of electrons from one mole of Li atoms on the surface of Li metal). Electrons are emitted from the Li metal surface when the metal surface is illuminated with ultraviolet light of wavelength 245 nm. What is the kinetic energy of a photoelectron emitted from the surface of the metal when the ultraviolet light shines on it? (Hint: Use the binding energy of one electron)

Answer 1

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S01:1 photon let lithium threshold energy to Binding energy = do from photo electric effect, we know The = d.+ (1. El election wheree =) ( = Velocity of light = 3x108 mis h= plancit's constant 6.626x10-34 Js Azwanclength of photon; here given Binding energy = 279.7 kJ / mol = 6.022X1023 electron (we knows Imole Hence; Binding energy per election = 297.781000 S 6.022 % 1023 Tlelection - 4.95 X 10-9193 Also given; wavelength of nence Energy Incident photon = 245nm of photon hc = 6.626 X10-34 X 3X108 245x109 upon solving; we get - 8.11X105 Energy of photon 8-11 810-19 7 Hence; (K.E) Energy of photon - Binding Energy election - 8. 118 10-19] - 4.95 x 10-195 & 3.16810-195 Kinetic Energy of photoelectron emilled = 3.168 10 -19.5

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