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Homework Help: Photoelectric effect and blue light

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data

    Photoelectric effect
    A beam of blue light falls on the cathode of a photocell so that electrons are emitted. The blue beam is then replaced by a yellow one with the same intensity and electrons are also emitted. What would happen to each of the following physical quantities when the blue beam is replaced by the yellow beam?

    K: Maximum KE of electrons emitted
    I : The magnitude of photoelectric current

    A. K increases, I uncahged
    B. K decreases , I unchanged
    C. K decreases, I increases
    D. K decreases, I decreases

    2. Relevant equations

    Photoelectric equation : KE(max)= hf - (work function)

    h = planck constant, f = frequency of incident light, KE(max) = maximum KE of photoelectron

    3. The attempt at a solution

    Since yellow light is now used, frequency decreased --> so K decreases
    Since Intensity is unchanged, current unchanged --> so I unchanged
    so I chose B.

    Correct Answer : C

    Which part did I understand wrongly? Thanks !
  2. jcsd
  3. Sep 14, 2011 #2
    I is the magnitude of the photoelectric current, not the intensity of the incident beam as stated in the question

    since current = charge per unit time,

    if you have less electron ejected per unit time due to the lower frequency yellow light, your photoelectric current magnitude will thus decrease

    hmm... maybe the answer is D?

    i don't know... someone else might know
    Last edited: Sep 14, 2011
  4. Sep 14, 2011 #3
    but since (KE)max decreases, doesn't it mean that the current decreases?
  5. Sep 14, 2011 #4
    sorry, i was still editing my answer

    but , anyway, i don't think its because KE decreases, that current decreases

    its more about the frequency of photon ejecting the electrons that will determine the current magnitude

    current is charge per unit time.

    if you switch to yellow which is lower frequency, photons will thus have a lower frequency of hitting out the electrons, so the current should decrease.
  6. Sep 14, 2011 #5
    Light intensity tells us how much energy is delivered per second.

    To deliver the same energy per second with low energy yellow photons means more photons per second

    If we assume every photon releases an electron then there are more electrons emmitted per second

    I increases. C is correct.
  7. Sep 15, 2011 #6
    hmm , if you say more photons per second, wouldn't that mean the frequency of yellow light increase?

    if i put it mathematically,

    I = P/A , P = E/t
    I = E/ (tA) , E = hf
    I = hf / (tA)

    if i switch to yellow light, frequency would drop. so for constant I, A has to drop?
    since area drops, you have lesser emitted photoelectrons for a smaller area per unit time, and hence current should drop?

    but also from wiki,
    so since the yellow incident light has the same intensity as blue, wouldn't it then mean that the rate of photoelectrons ( which is current) is the same? so magnitude of photoelectric current should not change?
  8. Sep 15, 2011 #7


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    E = hf is the energy of a single photon, not the energy of the entire beam. To compute the intensity this way, you would need to know, and account for, the number of photons incident on the area per unit time. This won't be the same for different frequencies and intensities.
    Last edited: Sep 15, 2011
  9. Sep 15, 2011 #8


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    This means for a fixed frequency, i.e. for the same colour light. In the case of the above experiment, the intensity is fixed, while the frequency is varied.

    apelling is correct.
  10. Sep 15, 2011 #9
    doesn't the frequency of light , e.g yellow, tell you the number of incident photons per unit time?

    so if intensity is fixed, then like what apelling said, that more photons per sec are required for yellow light to make intensity constant, then wouldn't it mean that yellow light has increased frequency which makes it a blue light now?
  11. Sep 15, 2011 #10
    No the rate at which photons arrive is not the same as the frequency of the light.
    The frequency of the light is the frequency of oscillation of the electromagnetic fields within each photon wave packet. This is the frequency in E=hf. The number of photons per second is indeed proportional to the intensity but inversely proportional to the frequency of the light.
  12. Sep 15, 2011 #11


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    Just to clarify what apelling said to avoid further confusion:
    • The flux of photons per unit time is proportional to the intensity for fixed frequency
    • The flux of photons per unit time is inversely proportional to the frequency for fixed intensity
  13. Sep 15, 2011 #12
    wow i never knew these... :X

    thanks everyone!
  14. Sep 15, 2011 #13
    Thanks for all the help ! I understand it clearly now !
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