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Help with an exam question

  1. Jun 28, 2005 #1
    Electromagnetic radiation of wavelength 4.50 x 10^-7 m and intensity 800Wm^-1 is incident on the surface of a metal with work function of 2.02eV. Electrons are emmited from the surface and attracted to a positively charged electrode with a p.d. of 10V.

    Question: Assuming that each photon liberates an electron, estimate the pressure exerted by the photoelectrons on the positive lectrode which has an area of 3.0 x 10^-4m^2. ( You may assume that all the photoelectrons emmited are absorbed by the positive electrode.)

    In the exam, I gave the answer in terms of distance, d, which is the distance between the metal and the positively charged electrode. Am I right to say that this question is unqorkable if no distance is provided?
     
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  3. Jun 28, 2005 #2

    OlderDan

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    I assume you calculated an initial energy for the emitted electrons based on excess photon energy relative to the work function. The increase in energy depends only on the potential difference between the metal and the positive electrode. The distance does not matter. If the distance is shorter, the electric field will be stronger and accelerate the electrons in the shorter distance.
     
  4. Jun 29, 2005 #3
    I calculated the K.E. it had when the photoelectrons were ejected from the metal. But I needed the distance and same too for the electric field. Perhaps you could express your ideas mathematically so I can see where my error lies in.
     
  5. Jun 29, 2005 #4
    The electrons inial energy will be part kinetic and part potential energy. This potential energy will be converted to kinetic energy when th electron hits the electrode thus increasing the electrons velocity by an amount independent of the distance covered.

    [tex]\frac{1}{2}mv^2+eV=\frac{1}{2}mv'^2 \rightarrow v'=\sqrt{v^2+\frac{2eV}{m}}[/tex]

    This, i.e. by using energy conservation, is the easiest way to solve the problem. You could ofcourse also do it by solving the equation of motion for the electron, but this is a lot more work and will yield the same result:

    Assuming a homogeneous electric field V/d this yields a constant acceleration a

    [tex]F=ma \rightarrow eV/d=m \frac{d v}{dt} \rightarrow v'=v+a t[/tex] with [tex]a= \frac{eV}{md}[/tex]

    With the relation for constant acceleration

    [tex] d=vt+\frac{1}{2}a t^2 \rightarrow t=-v/a+\sqrt{\frac{2d}{a}+{v^2}{a^2}}[/tex]

    Inserting this in the equation for v'

    [tex]v'=v+a t=v+a(-v/a+\sqrt{\frac{2d}{a}+\frac{v^2}{a^2}})=v+(-v+\sqrt{2da+v^2})=\sqrt{v^2+2da}=\sqrt{v^2+\frac{2deV}{md}}[/tex]

    where you can see d drops out and the same result as with energy conservation is obtained.
     
  6. Jun 29, 2005 #5
    Hmm, but the question asks for the pressure felt by the positive electrode. That means I must calculate the force acting on the positive electrode. Wouldnt I have to know the distance between the metal and the positive electrode?
     
  7. Jun 29, 2005 #6

    Doc Al

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    All you need to know is the speed of the electrons when they hit the electrode; the distance the electrons had to travel is irrelevant.

    What force must the electrode exert on the incoming electrons to bring them to rest? Make use of the fact that impulse equals the change in momentum: [itex]F \Delta t = \Delta (mv)[/itex].
     
  8. Jun 29, 2005 #7
    i see... Butda_willem gave time as [tex] d=vt+\frac{1}{2}a t^2 \rightarrow t=-v/a+\sqrt{\frac{2d}{a}+{v^2}{a^2}}[/tex]

    , dont i have to use distance?
     
  9. Jun 29, 2005 #8

    Doc Al

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    Look at da_willem's entire post. He shows that the distance drops out of the final answer, no matter which method you use.
     
  10. Jun 29, 2005 #9
    I still dont get it :frown:

    da willem found out the final velocity when the photoelectrons hit the positive electrode and only proved that d was dropeed out of the expression for final velocity. So to find out the force acting on the positive electrode,

    [tex] m_e(\frac{v-u}{t}) [/tex]

    And since da willem proved that [tex] \sqrt{v^2+\frac{2deV}{m_ed}} [/tex],

    I need the time, so how can I get rid of the d in the expression for time?
     
    Last edited: Jun 29, 2005
  11. Jun 29, 2005 #10

    Doc Al

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    Realize that the [itex]\Delta t[/itex] in the impulse equation has nothing to do with the transit time of the electrons.

    Think this way: How many electrons hit the electrode per second? What is their change of momentum?
     
  12. Jun 29, 2005 #11
    Ok, so force is the change of linear momentum divided by the change in time. To calculate the force acted by a single photoelectron, wouldnt I have to know the acceleration of the particle at the instant and thus the transit time? Please help me, I just dont understand how the transit time is not needed
     
  13. Jun 29, 2005 #12

    Doc Al

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    You are thinking of the time it takes for a single electron to be brought to rest as it hits the electrode. This has nothing to do with the time it takes for the electron to get from the surface to the electrode. (Assume that the electrons are stopped quickly.)

    Rather that worry about the dynamics of a single electron, calculate the average force by estimating the number of electrons that must be stopped per second. Then use [itex]F = \Delta (mv)/\Delta t[/itex].
     
  14. Jun 29, 2005 #13
    If the electronswere accelerating at that moment when they hit the positive electrode, arent you assuming that the electrons reach the positive electrode with a constant velocity, and the deceleration is just [tex] \frac{v}{t} [/tex]?
     
  15. Jun 29, 2005 #14

    Doc Al

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    I do assume that the electrons have some speed v when they reach the electrode. (Which you can calculate in any of several ways.) But I have no idea what "t" (the time it takes to stop an electron) is so I can't directly calculate the deceleration of each electron. (That "t" is not the travel time of the electron from surface to electrode!) But if I know the number of electrons emitted per second, I can calculate the average force.
     
  16. Jun 29, 2005 #15
    I dont see how you can calculate the average force by just knwing the number of electrons that are emmited. Perhaps you can express your ideas mathematicaliy so i can understand better.
     
  17. Jun 29, 2005 #16

    OlderDan

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    You know the intensity of the radiation. You have calculated the energy per photon, [tex] E_p [/tex]. The rate at which photoelectrons are emitted is also the rate at which they strike the positive electrode. In a time [tex] \Delta t [/tex] the number of electrons that strike the electrode is determined by the rate at which the energy is absorbed by the photoelectrode and the energy per photon:

    [tex] \Delta N = I \cdot \frac{Area}{ E_p }\Delta t [/tex]

    where I has units of

    [tex] = \left[ \frac{Watts}{m^2} \right] [/tex]

    You need to check your original problem statement to get the units of intensity correct. You will have to assume the area of the photoelectric material is the same as that of the electrode unless you have been given a different area.

    Each photoelectron has velocity v when it strikes the electrode, so it has momentum mv which it loses in the collision. The electrode provides the force that changes the electron momentum, and by Newton's third law the electron exerts an equal and opposite force on the electrode. You cannot know how much that force is and you don't need to know. What you need to find is the average force applied to the electrode, which is the total momentum lost by electrons per time interval [tex] \Delta t [/tex]. The total electron momentum lost in time interval [tex] \Delta t [/tex] is

    [tex] \Delta P = mv \Delta N = mvI \cdot \frac{Area}{ E_p } \Delta t[/tex]

    Can you take it from there?
     
  18. Jun 30, 2005 #17
    I know what you guys are trying to say. The problem, lies in that the photoelectrons have an acceleration when they hit the positive electrode. Your equations didnt take into account this acceleration, thats why I cant understand why the distance isnt needed.
     
  19. Jun 30, 2005 #18

    OlderDan

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    The photoelectrons accelerate from the photoelectric material where they are emitted toward the positve electrode. That acceleration increases their velocity by an amount that is completely determined by the 10 V potential difference as I said in my forst post and as others have shown. That accelertion and the consequent velocity of the electrons striking the electrode has been accounted for. You do not need to calculate that acceleration. The final velocity of the electrons is calculated using conservation of energy, and the distance the electrons travel while the electrical potential energy is being converted to kinetic energy does not matter. All that matters is the potential difference.

    If you are talking about the deceleration of the electrons when they hit the electrode, that is incorporated in the momentum change of the electrons. The details of the deceleration depend on the exact nature of the force acting on each electron. You cannot know the details of the process whereby these electrons are captured by the electrode, so you will never be able to find the deceleration. You do not need to know it. All you need to know is the change in momentum of each electron and how many electrons hit the electrode per time interval to calculate the average force.

    If this is the part you are stuck on, you need to review the relationships between force, impulse, and momentum. Impulse is the integral of force over time, or the average force acting for some time interval times the length of that interval. You cannot know exactly how the force varies with time, and it might very well be different for different electons hitting the electrode. No matter how the force varies in time, the impulse of the force on each electron is going to be equal to the change in momentum of that electron. You cannot find the force on an individual electron, but you can find the average force over all the electrons that hit the electrode over some interval of time. That is all you can do, and it is all you have been asked to do.
     
  20. Jun 30, 2005 #19
    I know what you are trying to say. You are trying to say that what I am suypposed to find is Force/s acting on the positive electrode. I know the whole picture of the average force thingy.

    What Im still clueless about is how you didnt take into account the acceleration of the electrons when they hit the positive electrode. Lets say if the electrons reach the positive electrode, it would have an acceleration, wouldnt the force exerted by the positive electrode have to be larger to counter the acceleration and prodeuce a decelerating force?

    Wouldnt the change in momentum of the electrons be larger than simply [tex] m_ev_f [/tex]? If the electrons were accelerating, wouldnt the momentum of each electron be increasing as well if not for a decelerating force?

    Would be glad if you could clear my misconceptions. I really appreeciate the help you all are rendering me right now.
     
  21. Jun 30, 2005 #20
    Thank you all for all your help. Ive finally realised where I had gone wrong. When the photoelectrons reach the positive electrode, they dont have an acceleration since the potential is used in doing work for bringing the electron towards the positive electrode.

    Thanks for your patience, I realy appreciate the help.
     
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