Light intensity, Max KE of photoelectrons

[AznBoi]

Please check to see if my solution is correct:

Question:
If a light source were increased by a factor of 2 in intensity, what would happen to the value of KE_max of each photoelectron?

My solution:
KE_max would stay the same because increasing the intesity by a factor of two would only increase the amount of photoelectrons by a factor of 2. However each photoelectron would still have the same amount of energy KE=hf-\phi since each electron only absorbs energy from one photon.

Is my solution entirely correct?

 

[turdferguson]

Looks good

 

[AznBoi]

Okay thanks!

 

[AznBoi]

So for any EM source, to calculate it's wavelength you would always use this equation: v=f*\lambda right? Do you always use the speed of light (c=3x10^8) for (v) if the radiation is electromagnetic? It would only change if the wave was propagated through another medium right?

 

[cepheid]

So for any EM source, to calculate it's wavelength you would always use this equation: v=f*\lambda right?

Yes, this is basic:

v = \frac{\lambda}{T}

where T is the period. This equation says that the speed of the wave can be calculated from the distance it travels in one period divided by the duration of one period (speed = distance / time, after all). Of course, 1/T = f. That's how you can easily understand and recall this equation.

Do you always use the speed of light (c=3x10^8) for (v) if the radiation is electromagnetic?

No.

It would only change if the wave was propagated through another medium right?

Yes. (The answer to this question explains the answer to the previous question. The two assertions cannot both be true, of course. Either the speed changes or it is always c. 'Always' is an absolute).

 

[AznBoi]

Thanks for allowing me to comprehend all of this! I understand it now. =]

Also I'm confused about this:
Relate the linear momentum of a photon to its energy or wavelength , and apply linear momentum conservation to simple processes involving the emission, absorption, or reflection of photons.

p=mv \qquad I know that \quad E=KE=\frac{1}{2}mv^{2}=hf-\phi = h\frac{v}{\lambda}-\phi but I still don't get exactly what they are asking, how do you combine these two if the momentum is missing the 1/2 and square? (^2)? I don't get the rest of the objective either.

Also, following Thompson's cathode ray experiment, how do you derive the charge to mass ratio? I know that but I'm confused on how to simply that further to get: q/m= _____. Please help, Thanks!

 

[cepheid]

Hmm...well keep in mind that photons are massless particles, and so the relation p = mv is not the correct relation for the momentum. I think the correct relation comes from special relativity and is given by:

E = pc

==> p = E/c = h/lambda

I'm not sure whether you are expected to know this at this stage in your studies.

 

[AznBoi]

Oh yeah, that's on my equation list. Okay, thanks now I know what to look for. =]

 

[AznBoi]

Wow, so far I've gotten many of these atomic problems wrong because of the units. When should I convert (eV) to Joules of vice versa? It's hard to determine when you need to convert and when you don't. =/

 

[cepheid]

Hmm...well keep in mind that you can't go wrong if *everything* is in SI units, in which case you'd measure energy in Joules. But I can't think of a general set of circumstances in which it's more convenient to keep things in eVs. I'd have to see a specific example.

 

[AznBoi]

Okay, I think I'm just going to convert everything to SI units because if you leave (eV) units in there, you aren't able to use (kg) or (m) units right? Thanks for all of your help!

 

[turdferguson]

In order to use 6.6*10-34 for Planks constant, things need to be in Joules. To work in eV for things like energy levels, use 4.1*10-15 eV*s