Achieve Homework #2, Question #13

[trishaferrer]

Hi! Can someone help me with this? Thanks!

[205484435]

Hi,

For the first part of the question, we are trying to find the E per photon which is E(photon) = hv
we are given that the wavelength is 2.6 um or 2.6*10^-6 m, given that we can find the frequency using v = c/lambda and then plug in v into E(photon) = hv or you can just use E = hc/lambda

[205484435]

For the second part of the question you are trying to find the energy of the electron which is the kinetic energy : Ek = 1/2*m*v^2
so we need the velocity of the electron which you can find using deBroglies equation and rearrange it to find velocity = h/m*lambda (we are using the same wavelength given 2.6*10^-6m) once you find the velocity you can just plug it in and we already know the mass of an electron which is 9.109*10^-31kg.

[carrie_karchmer_3D]

I solved the first part of this problem using E= hc/wavelength, this will give you the energy of the photon. Make sure to convert the wavelength units to m. For the second part of the problem, I used de Broglie to solve for velocity. Once I had the velocity, I used KE = 1/2mv^2 to find the energy of the electron.

[Amanda Tran 1D]

Remember that when finding the energy of a photon, you can use the equation E = hv, but when finding the kinetic energy of an electron, you should use the equation E = (1/2)mv^2.

[Hope Galarneau 1A]

For the first part of the question, you would use E=hv since its asking for the energy of the photon. For the second part, since its asking for the energy of the electron, you would use E=1/2mv^2. Also a good way to differentiate which equation you should use is by looking at what it's asking you for, so since the second part is asking for the energy of an electron, and E=1/2mv^2 uses the mass of an electron, you could assume that this would be the correct equation.

[Alexis Rempillo 3B]

For the second part, I also suggest lambda_e- = h/p = h/(mv) to find v. I tried using c = lambda(v), but I got an incorrect answer.