What Are the Formulas for Calculating Photon Momentum and Work Function?

  • Thread starter Thread starter Procrastinate
  • Start date Start date
  • Tags Tags
    Photoelectric
AI Thread Summary
Binding energy is often considered equivalent to work in the context of photon interactions. For a He-Ne laser with a wavelength of 6.3 x 10^-7 meters, the change in momentum of each photon upon reflection can be calculated using the correct formula for photon momentum, which is not p = mv. The discussion highlights confusion around the mass of a photon, emphasizing that photons are massless and should not use the mass of an electron for calculations. The work function is clarified as the energy required to release an electron, distinct from general work. Understanding the correct formulas for photon momentum and the work function is essential for solving these physics problems.
Procrastinate
Messages
155
Reaction score
0
1. Is binding energy equivalent to work (W)?

2. A He-Ne gas laser produces monochromatic red ligth with a wavelength of 6.3 x 10-7. If the beam is reflected normally from a mirror, what is the change in momentum of each photon on reflection?

I am particularly confused about this question. I used delta p = m(v - u) where v was the speed of light but it didn't result in the correct answer. Could anyone please give me a hint?

I also couldn't figure out what the mass of a photon was so I used the mass of an electron which I am sure is wrong.
 
Physics news on Phys.org
Procrastinate said:
1. Is binding energy equivalent to work (W)?

Is this work, or "work function"?

2. A He-Ne gas laser produces monochromatic red ligth with a wavelength of 6.3 x 10-7. If the beam is reflected normally from a mirror, what is the change in momentum of each photon on reflection?

I am particularly confused about this question. I used delta p = m(v - u) where v was the speed of light but it didn't result in the correct answer. Could anyone please give me a hint?

I also couldn't figure out what the mass of a photon was so I used the mass of an electron which I am sure is wrong.

You need to go back and find out the expression for the momentum of light. Hint: it is NOT p=mv!

Zz.
 
ZapperZ said:
Is this work, or "work function"?

Work function: E = W + KE



You need to go back and find out the expression for the momentum of light. Hint: it is NOT p=mv!

Zz.

1. I supposed binding energy many the threshold energy that was required to illuminate light.

2. Is it de Broglie's wavelength?: p = c/lambda

I didn't think it was this because the change in momentum is a different formula.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top