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# Question on the photoelectric effect. Watch

1. I don't normally get stuck on that many summary questions, but I can't seem to answer question 2)b) on page 31 in the new AQA Physics A textbook. The answer to this question isn't in the back of the book and I was hoping someone on here would be able to explain it to me! :)

The question goes:
"A metal surface at zero potential emits electrons from its surface if light of wavelength 450nm is directed at it. However, electrons are not emitted when light of wavelength 650nm is used. Explain these observations."

My thought process:
I assumed it had something to do with the threshold frequency and work function of the metal. My first instinct would be to work out the energy of the photons (E=hf) and to compare. But, I get stuck at this point, as surely a photon can't have too much energy (this would be converted to kinetic energy upon photoemission of electron) or too high a wavelength (as long as it exceeds the threshold frequency.

I also assumed c=f x lambda would come in here somewhere.
2. n.b. I misused "photon" and "electron" in here a few times, which I have just noticed. This question has me confused for some reason
3. (Original post by ryanaue)
I don't normally get stuck on that many summary questions, but I can't seem to answer question 2)b) on page 31 in the new AQA Physics A textbook. The answer to this question isn't in the back of the book and I was hoping someone on here would be able to explain it to me!

The question goes:
"A metal surface at zero potential emits electrons from its surface if light of wavelength 450nm is directed at it. However, electrons are not emitted when light of wavelength 650nm is used. Explain these observations."

My thought process:
I assumed it had something to do with the threshold frequency and work function of the metal. My first instinct would be to work out the energy of the photons (E=hf) and to compare. But, I get stuck at this point, as surely a photon can't have too much energy (this would be converted to kinetic energy upon photoemission of electron) or too high a wavelength (as long as it exceeds the threshold frequency.

I also assumed c=f x lambda would come in here somewhere.
Try using c = f * wavelength in E = hf
4. (Original post by rcmehta)
Try using c = f * wavelength in E = hf
Ahhh I see where I went wrong now, I went straight to E=hf and forgot that the question gave me wavelength and not frequency, haha. Thanks for clearing that up, blonde moment
5. (Original post by ryanaue)
Ahhh I see where I went wrong now, I went straight to E=hf and forgot that the question gave me wavelength and not frequency, haha. Thanks for clearing that up, blonde moment
Happened to me enough times xD, no problem

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Updated: April 10, 2016
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