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### Frday's lecture

Posted: Sat Jun 29, 2019 2:17 pm
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?

### Re: Frday's lecture

Posted: Sat Jun 29, 2019 2:51 pm
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.

### Re: Frday's lecture

Posted: Sat Jun 29, 2019 7:42 pm
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?

### Re: Frday's lecture

Posted: Sun Jun 30, 2019 11:36 pm
Thank you for relating both equations step by step, my brain appreciates it!

### Re: Frday's lecture

Posted: Mon Jul 01, 2019 8:20 am
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)