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.
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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.
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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 \(c\).
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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,
\(v=\dfrac{\Delta d}{\Delta t}=\dfrac{~~~\lambda~~}{\frac1f}=\lambda f=c\)
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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 \(E=h\dfrac c\lambda\).