## Homework 1.33 Part C

$\lambda=\frac{h}{p}$

Alicia Yu 1A
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### Homework 1.33 Part C

Part C asks "what is the wavelength of the radiation that caused photoejection of the electron?" Is that referring to the wavelength of the electron that is emitted, or is it talking about the energy that is needed to eject the electron? Could you just use the E=hc/wavelength formula or would you have to include kinetic energy?

Chem_Mod
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### Re: Homework 1.33 Part C

The radiation it is referring to is electromagnetic radiation, which is light or a photon. The photon "caused photoejection of the electron." An electron is not considered a type of "radiation." We can use $E=\frac{hc}{\lambda}$, but the energy of the photon was responsible for both the separation of the electron from the metal and the kinetic energy of the electron. We take this into account using $E=\frac{hc}{\lambda}=\phi + E_{k}$

Tiffany Chen 1A
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### Re: Homework 1.33 Part C

So kinetic energy is this case would be $\frac{1}{2}mV^{2}$ and we would just plug in the mass of an electron and the given velocity?

Also, how would you be able to tell whether or not you should account for kinetic energy? Do we include kinetic E here, because we know that this problem is about the photoelectric effect?

Cytlalli 1B
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### Re: Homework 1.33 Part C

I believe that going back to Friday's lecture we talked about how energy is released through photoejection of an electron meaning there would be kinetic energy. So overall, yes, we know to use kinetic energy because it is about the photoelectric effect. Photoelectric problems follow the format of this equation: E(photon)- Φ= E(kinetic) or some variation of this equation.