# Kaplan High Yield Problems Physics - Photoelectric Effect (p428)

Discussion in 'MCAT Study Question Q&A' started by mejorization, Jul 28, 2011.

1. ### mejorization 7+ Year Member

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When a photon hits a metal surface, it liberates an electron. Thus, the number of photons is equal to the number of electrons liberated. After the electron is liberated, it should have kinetic energy left over (since photon energy-work function=electron energy). Does this leftover kinetic energy affect the current at all?

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2. ### Majik

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Hmm. From what I can recall (don't take my word on this) but I believe the photoelectric effect is something like this:

E = work function + KE, where E is basically the ionization energy needed to eject an electron off the atom. Any energy in excess of the work function is utilized as Kinetic Energy of the electron. Anyways, with regard to your question, I'm not really sure how it pertains to current flow since I've never encountered an example where I saw this, but from my understanding, regardless of what kinetic energy an electron has, no current can be produced without a voltage source (potential difference). In the absence of a potential difference, the electrons randomly collide into the walls of a wire for example, but there is no net ("drift velocity"), therefore there is no current.

3. ### ih2183

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I think theres a correlation between light intensity and current produced by freed electrons. KE is proportional to hv (frequency), but frequency and intensity are two different concepts. KE for a given frequency is fixed, but it is really intensity of a given frequency that affects the current produced, I believe.

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### mejorization 7+ Year Member

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So basically, higher kinetic energy will not result in higher current? I guess that means the kinetic energy is lost as heat.

Got it. Higher intensity=higher current. Higher frequency=able to displace materials with higher work function. Sound about right?

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5. ### ih2183

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work function is a constant that varies with the metal. Different metals require different hv to overcome work function.

6. ### warypremed 2+ Year Member

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Can anyone help me with the similar question #3?

What power and frequency of incident radiation must be used to strike a metal (of work function 1 x 10^-18 J) to produce 10,000 electrons per second?

I got the frequency but I don't know how I'm supposed to get to power.

7. ### Bumbl3b33 Removed

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Power = work/time

8. ### warypremed 2+ Year Member

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Wow did not realize how easy this was. One simple multiplication. Thanks

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