Photoelectric Effect

[Dart82]

I am having trouble figuring how the info given relates to electric and magnetic fields.

1. The problem statement, all variables and given/known data
A laser emits 1.38x10^18 photons per second in a beam of light that has a diameter of 1.96 mm and a wavelength of 518.0 nm. Determine each of the following for the electromagnetic wave that constitutes the beam.
(a) the average electric field strength in N/C

(b) the average magnetic field strength in T



2. Relevant equations
E = F/q for electric field

F = qvB sin(theta) for magnetic field

E=hf=hc/lambda for energy of a photon



3. The attempt at a solution
i solved for E the energy of a photon:
h*c/(518x10^-9m) = 3.84x10-19 Joules

I am not just some slacker begging for an answer; i have a high B in this class! i don't see how in the world the concept of the photoelectric effect can relate to electric and magnetic fields if i don't know q, F, B. I am completely lost here - someone please help!!!

[Dick]

The photoelectric effect is important here because you are given the rate of emission of photons in the beam and the wavelength (hence energy) of each photon. Together with dimensions of the beam this gives you the energy density in the beam. This energy density could also be calculated from averaged electric and magnetic fields associated with the light beam. So knowing the energy density should let you calculate the electric and magnetic field energy - hence the field intensity.

[Dart82]

Do i add the area of the laser beam to the energy of each photon (E)? this is so frustrating....

[Dick]

Do i add the area of the laser beam to the energy of each photon (E)? this is so frustrating....

How can you add an area to an energy??? Look, the laser runs for one second, emits 1.38x10^18 photons. Those photons are in a cylinder 1.96 mm in diameter and c*1sec long. What's the energy density?

[Dart82]

How can you add an area to an energy??? Look, the laser runs for one second, emits 1.38x10^18 photons. Those photons are in a cylinder 1.96 mm in diameter and c*1sec long. What's the energy density?
sorry, i misunderstood what you had said about the dimensions of the beam. i think i need to take a break and try this one again later. 5 hours of Physics homework has tweaked my brain a bit too much. thanks for your help Dick.

[Dick]

Yeah, take a break. BTW, this isn't a photoelectric effect problem either.

[Dart82]

got it... no photoelectric effect.