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Photoelectricity

Doc Boff Physics AS level English 14 cards

Photoelectricity

Cards in this deck (14)

  1. 1Photocell diagram:Photocell diagram:
  2. 2How would we use a photocell?Shine UV light on cathode so electrons released, which are attracted to the anode, causing current to flow
  3. 31 electron is liberated by.....1 photon
  4. 4Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz What is the velocity of the electrons leaving the plate?(2.0 x 10^15) x (6.63 x 10^-34) =
    1.33 x 10^-18 J
    (1.33 x 10^-18 ) - (6.88 x 10^-19) =
    6.42 x 10^-19 J
    Root(6.42 x 10^-19 / 0.5 x (9.11x10^-31)) = 1.2x10^6 ms-1
  5. 5Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz How many photons are hitting the plate per second?(2.0 x 10^15) / (1.60 10^-19) = 1.25x10^7 electrons
  6. 6Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz What would happen if the light frequency doubled but power remained constant?1/2 number of electrons ejected per second with a greater Ek, therefore the current halves
  7. 7Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz What would happen if a second identical lamp is added at the same distance?Double number of photons hitting plate per second, therefore double electrons ejected, Ek is the same, so current doubles
  8. 8Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz What would happen if the surface area of the zinc plate was halved?Number of photons hitting plate per second halves, and so does the number of electrons ejected per second, Ek is the same, so current halves
  9. 9Photocell contains zinc with ∅ = 6.88x10^-19 J Original current = 2.0x10^-12 A Frequency of light = 2.0 x 10^15 Hz What would happen if the original lamp was moved back 3 times as far?9 times fewer electrons ejected per second, so 9 times fewer photons hit plate per second, Ek is the same, so current is divided by 9
  10. 10Electrons that leave the plate on a photocell can be attracted back with....Sufficient positive charge
  11. 11What is stopping potential?The potential difference necessary to stop all electrons (even the electron with the most Ek) from reaching the opposite plate
  12. 12What is the equation for stopping potential?Ek(max) = eVs
    (e = elementary charge)
    (Vs = stopping potential)
  13. 13If the threshold frequency is not met, what effect does increasing the intensity of the light have?No effect at all
  14. 14If the threshold frequency is met, what effect does increasing the intensity of light have?More electrons released per second, so more photons striking plate per second

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Photoelectricity
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