## Week 2,3,3 Sampling #25

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

VanessaZhu2L
Posts: 93
Joined: Wed Sep 30, 2020 9:44 pm

### Week 2,3,3 Sampling #25

For 25, when finding the energy of the photon and electron, the solution gives two different ways to solve it.

For the photon, they use E=hv and c=vλ and for electron, they used λ=h/p and e=1/2mv^2.

I was wondering why they used the two different methods and how you know which equations work with the electrons or the photons.

Thank you!

Posts: 159
Joined: Wed Sep 30, 2020 9:32 pm
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### Re: Week 2,3,3 Sampling #25

The equations E=hv and c=vλ refer to electromagnetic radiation and therefore can only be used for photons. On the other hand, λ=h/p and ke=1/2mv^2 are for any particles that have rest mass. The second two equations can be used for electrons, neutrons, protons, etc. Since photons have no rest mass, the DeBroglie and kinetic energy equations cannot be used for them.

Hope this helps!

AnnaNovoselov1G
Posts: 78
Joined: Wed Sep 30, 2020 9:51 pm
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### Re: Week 2,3,3 Sampling #25

I also wanted to add that you can use c=vλ and E=hv when calculating whether an electron can be ejected because electrons absorb the energy of a photon that strikes them. For instance, an energy of light (hv) that equals the work threshold of a metal could eject the electrons (= the energy the electrons would get). But directly, E=hv and c=vλ can only be used for photons because they are specific to electromagnetic radiation.