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historygeek96
Light with a wavelength of 5.0 · 10-7 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.
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:dw|So 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 \(E=h\dfrac c\lambda\).
Do you have any questions?
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.
Haha, okay! If not, feel free to let me know.