## Frday's lecture

$c=\lambda v$

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Karolina herrera1F
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Joined: Sat Feb 02, 2019 12:15 am

### Frday's lecture

Im still not sure why if c=lamba v and v=c/lamba give you the equation E= hv= hc/lamba. Is it because you combine both of them, but why would you combine them?can someone explain this to me?

Jeril Joseph 1B
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### Re: Frday's lecture

That's basically the reason. You substitute the v for the c/lambda, and because that's multiplied by h, you put them together to make E=hc/lambda. Then, if you want to solve for the wavelength, you cross multiply the equation, and divide both sides by e to get lambda=hc/e.

Karina Vasquez 1D
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Joined: Fri Mar 01, 2019 12:15 am

### Re: Frday's lecture

I know this doesn't really have to do with the question, but how do you find the threshold energy, or would it always be given? In a vague way? So for example, in the second question from the lecture I'm assuming the threshold is 3.61 x 10^-19 J. Is that going to be the threshold unless it explicitly states that the threshold energy is something different?

Kguox1B
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Joined: Mon Jun 17, 2019 7:22 am

### Re: Frday's lecture

Thank you for relating both equations step by step, my brain appreciates it!

Chem_Mod
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### Re: Frday's lecture

Karina,

The threshold energy could be directly provided, such as the "threshold energy is...". Or it can be implied such as in the lecture example, "An X amount of energy is applied and the electron had no kinetic energy" which means you had enough energy to only release the electron, which means you only applied the necessary amount- the threshold energy.

A question may provide an assortment of info, such as velocity, wavelength, frequency, and momentum and it may want you to find the threshold energy.

An important part for success of the photoelectric effect section is really understanding all the different things you can find from the E(photon)=E(threshold)+E(kinetic)

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