Explanation on "Doubling the Momentum of a Photon"?

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CarefreeSurfboard

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Hi, this question comes from the free Kaplan CBT Diagnostic Exam.

Passage:
http://i.imgur.com/74dQ6Dk.png

Question:
VH2FyLC.png


My reasoning for choosing D at first stemmed from the equation for momentum: p = mv. Seeing as the passage addressed the "discrete packages" of light that support the Photon Theory of Light, I assumed that p = mv was relevant since light was assumed to have some mass-like characteristics rather than purely wavelike characteristics.

But I'm dumb, I just thought going twice the speed of light sounded cool. 😎 (and photons of light don't have varying mass? not sure.)

Anyway, can someone clarify what the explanation means by "If the photon has twice the momentum, it has twice the energy."

  • I get that the answer involves the equations E = hf and c= fλ. If energy doubles, (h constant) then f doubles. If f doubles, then λ must halve (since c is constant).
Please help! Thank you very much. 😳

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I'm thinking

p = mv

If momentum doubles the Energy doubles because photons are all traveling at c and that isn't going to double

Therefore for momentum to double mass must double

And KE = 1/2mv^2
So energy of the system will double

Edit:
Ignore, didn't realize it was a photon, see other post
 
Last edited:
If momentum doubles the Energy doubles because photons are all traveling at c and that isn't going to double

Therefore for momentum to double mass must double

Are you saying that a "certain photon" can have a different mass from "another photon"?

And KE = 1/2mv^2
So energy of the system will double

I don't think kinetic energy was the aim of the addressed topic, but it does explain the mass of the photon doubling. Can anyone confirm on this?

I thought the question was more concerned with frequency and wavelength which is why I mentioned E=hf as well as c=fλ. It explains the answer. Doesn't explain the mass part though.
 
Actually now that I think about it, I take my thing back

I'm not sure if a photon even has mass


It's just

E = hc/lambda this is from E = hf and v = f*lambda so f = v/lambda

If lambda halves, E will double

My bad the first time
 
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Actually now that I think about it, I take my thing back

I'm not sure if a photon even has mass


It's just

E = hc/lambda this is from E = hf and v = f*lambda so f = v/lambda

If lambda halves, E will double

My bad the first time

Photons have no mass. They're pure electromagnetic energy. This question is tricky for sure. Use the equation provided in the passage as-is (p=E/c) to prevent confusion. If the momentum, p, increases by a factor of 2, then either E doubles, or c is cut in half. Since c is constant, E must double. For the E of a photon to double, it must either double it's frequency (not an answer choice), or halve its wavelength (from possibly the MCATs favorite equation, c=frequency•wavelength).

This is obviously easier to reason through after the fact, but it likely would cause many people problems, including myself.
 
Actually now that I think about it, I take my thing back
I'm not sure if a photon even has mass

It's just

E = hc/lambda this is from E = hf and v = f*lambda so f = v/lambda

If lambda halves, E will double

My bad the first time

No man, thanks for responding! Nothing to worry about. muchos gracias 🙂👍

Photons have no mass. They're pure electromagnetic energy. This question is tricky for sure. Use the equation provided in the passage as-is (p=E/c) to prevent confusion. If the momentum, p, increases by a factor of 2, then either E doubles, or c is cut in half. Since c is constant, E must double. For the E of a photon to double, it must either double it's frequency (not an answer choice), or halve its wavelength (from possibly the MCATs favorite equation, c=frequency•wavelength).

This is obviously easier to reason through after the fact, but it likely would cause many people problems, including myself.

From my understanding, the nature of light is a duality between mass and wavelike properties?

But apart from that, your answer is exactly what I'm looking for. Crystal clear explanation with the reference to the passage that I needed clarification on (p=E/c). Makes sense now. Thanks a lot!
 
No man, thanks for responding! Nothing to worry about. muchos gracias 🙂👍



From my understanding, the nature of light is a duality between mass and wavelike properties?

But apart from that, your answer is exactly what I'm looking for. Crystal clear explanation with the reference to the passage that I needed clarification on (p=E/c). Makes sense now. Thanks a lot!

Close. Not mass-wave duality, particle-wave duality. The current understanding is an amalgamation of the wave-like and particle-like theories (the wave-particle theory), but just remember that photons have no mass and are pure EM energy. Glad I could help!
 
Close. Not mass-wave duality, particle-wave duality. The current understanding is an amalgamation of the wave-like and particle-like theories (the wave-particle theory), but just remember that photons have no mass and are pure EM energy. Glad I could help!

Particle-wave duality (wave-particle theory) ... makes sense. Not to be nitpicky, but if photons are pure EM energy, how does that account for E=mc^2? Just as an electron has an extremely small mass, is it wrong to say that a photon has an extremely small mass?

I only ask because of this darn question... 😡
It seems as if the photon is "creating" particles (these being positrons and electrons). Or is it that the photon is a form of energy that is being converted into mass...

rTJGHN0.png
 
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