## De Broglie

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

Parth Mungra
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

### De Broglie

I was doing a problem in the 6th edition textbook (Chapter 1, 59), and I was confused on how to get energy per photon. It involved using the De Broglie equation, but I don't know how to obtain the correct units because they do not cancel at the end.

Jack Mitchell 3J
Posts: 24
Joined: Fri Sep 28, 2018 12:24 am

### Re: De Broglie

For that problem, do not use De Broglie's equation. Just use the equation Ep=hc/lamda because you are given the wavelength and not the momentum. Using this will the right answer and correct units.

Ian Marquez 2K
Posts: 66
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 2 times

### Re: De Broglie

Also, once you apply the equation E=hc/λ to find how much energy is produced by each individual photon, multiply the energy found from this equation with the total number of photons. This will cancel photons so your final answer will just be in just joules instead of joules/photon.