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historygeek96
Group Title
Light with a wavelength of 5.0 · 107 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency.
 5 months ago
 5 months ago
historygeek96 Group Title
Light with a wavelength of 5.0 · 107 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency.
 5 months ago
 5 months ago

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theEric Group TitleBest ResponseYou've already chosen the best response.1
Hi! Knowing wavelength (through air, I assume), you can find out the energy of the photons! The energy of a photon is always \(E=hf\). There, \(h\) is Planck's constant and \(f\) is the frequency. Many people use \(v\) for the frequency, too, as you'll see on Wikipedia. The frequency and wavelength are related by the speed. For a photon, the speed is the speed of light. In air, it's about \(3\times10^8\ \rm m\). We'll say that that speed is \(c\). _________________________________________________________ I'm going to show you how to relate these two, even though you only need the end result. Learn this if you can! It's handy to understand. Then we can use the velocity equation, where \(v\) is for velocity, and say \(v=\dfrac{\Delta d}{\Delta t}\)dw:1398209004730:dwSo distance is wavelength. The time per cycle (1 wavelength) is the reciprocal of the wavelengths per time. The cycles (wavelengths traversed) per time is the frequency. The reciprocal of the cycles per time is the reciprocal of frequency. Now, the displacement \(\Delta d\) is wavelength \(\lambda\). And the time per that displacement \(\Delta t\) is the reciprocal of frequency, so \(\dfrac1f\). Substituting, \(v=\dfrac{\Delta d}{\Delta t}=\dfrac{~~~\lambda~~}{\frac1f}=\lambda f=c\) ___________________________________________ Now \(c=\lambda f\implies f=\dfrac c\lambda\) Substituting that into what we had for energy, \(E=hf=h\dfrac c\lambda\) So, you can remember all that, or the \(c=\lambda f\), or the \(E=h\dfrac c\lambda\).
 5 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Do you have any questions?
 5 months ago

historygeek96 Group TitleBest ResponseYou've already chosen the best response.0
No i think that should do it. I never even know what formulas to use so now that I do that should be good enough.
 5 months ago

theEric Group TitleBest ResponseYou've already chosen the best response.1
Haha, okay! If not, feel free to let me know.
 5 months ago
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