Solving Physics Homework Questions: Kinetic Energy & Momentum of Photons

[ibysaiyan]

Homework Statement

Hi
I am not too sure but I had my physics exam the other day.I think it went great although I am much confused on few problems now when I come to think of them.Question first asked for me to find kinetic energy of a photon. So using 1/2mv^2 I got that value later on it asked me to find out the momentum of a photon.Now the wavelength was given to be around 660nm I plugged that into p = h/lamda. Now I think that's one way to do it or by having my K.e = value manipulating it such that mv = 2v times my k.e value. Now which is right or wrong :S ? :( Sorry but I just can't recall the values other than lamda which I posted above.
The next part of the question was quite complicated. They had Power of laser given as 1x10^-3 W and using force = change in momentum I had to find force.
So I used E (energy) = hc/lamda to get my energy value.Then by manipulation I got the time from power = energy/time. That time got plugged into force equation.
I hope I making any sense at all.
All inputs are greatly appreciated.
Thank you.

Homework Equations

The Attempt at a Solution

 

[americanforest]

The kinetic energy of a photon is not $$\frac{1}{2}mv^{2}$$. Plugging in the mass, m, of a photon as zero, we see that this classic formula for the kinetic energy returns zero, which is nonsense. The method using the de Broiglie wavelength is correct.

using force = change in momentum

You mean force = change in momentum per unit time.

$$\vec{F}=\frac{d\vec{p}}{dt}$$

Note that you used the formula $$E=pc$$ which is correct for particles with zero rest mass, such as photons. The general expression is

$$E^{2}=(pc)^{2}+(mc^{2})^{2}$$

You can save yourself some trouble by just using the equation for Power in terms of force and speed, $$P=Fv=Fc$$, where the last step is justified by the fact that photons move at the speed of light in vacuum.

 

[ibysaiyan]

The kinetic energy of a photon is not $$\frac{1}{2}mv^{2}$$. Plugging in the mass, m, of a photon as zero, we see that this classic formula for the kinetic energy returns zero, which is nonsense. The method using the de Broiglie wavelength is correct.


You mean force = change in momentum per unit time.

$$\vec{F}=\frac{d\vec{p}}{dt}$$

Note that you used the formula $$E=pc$$ which is correct for particles with zero rest mass, such as photons. The general expression is

$$E^{2}=(pc)^{2}+(mc^{2})^{2}$$

You can save yourself some trouble by just using the equation for Power in terms of force and speed, $$P=Fv=Fc$$, where the last step is justified by the fact that photons move at the speed of light in vacuum.

Err but in our specification we haven't met de Broiglies wavelength at all also the way the question was written down there clearly was mass and velocity(maybe it was an electron argh ) given now I just can't recall what those were but certainly if the both quantities are given then the way I did is right?
Yes force = delta momentum /time is what I used.
Now I just want assure myself on this.
Thanks for your reply =)

 

[americanforest]

Err but in our specification we haven't met de Broiglies wavelength at all also the way the question was written down there clearly was mass and velocity(maybe it was an electron argh ) given now I just can't recall what those were but certainly if the both quantities are given then the way I did is right?
Yes force = delta momentum /time is what I used.
Now I just want assure myself on this.
Thanks for your reply =)

The de Broiglie wavelength is simply $$\lambda=\frac{h}{p}$$, which is exactly what you said you used initially. If the problem was asking about an electron then one cannot use $$E=pc$$ since the mass is non-zero. The classical formula for kinetic energy remains relevant only for low momenta.

 

[ibysaiyan]

The de Broiglie wavelength is simply $$\lambda=\frac{h}{p}$$, which is exactly what you said you used initially. If the problem was asking about an electron then one cannot use $$E=pc$$ since the mass is non-zero. The classical formula for kinetic energy remains relevant only for low momenta.

Ah thanks a lot makes sense now. Seems I did right.
Thanks for your reply :)