Afternoon guys. Physics homework, stuck... nothing in my notes, or anyone else who seems to be able to help, so to you, Pit, I turn....

'Explain how the two statements below can each be justified by thinking of electromagnetic radiation as consisting of particles (photons).'

1- For any surface no photoemission occurs unless the frequency of the incident radiation is sufficiently high.

2- When the intensity of the incident radiation increases the maximum kinetic energy with which the electrons are emitted does NOT increase.

Thanks to anyone who can shed light... I'm also now very curious as to actually why, physics homework actually leading to curiosity! wow!
Afternoon guys. Physics homework, stuck... nothing in my notes, or anyone else who seems to be able to help, so to you, Pit, I turn....

'Explain how the two statements below can each be justified by thinking of electromagnetic radiation as consisting of particles (photons).'

1- For any surface no photoemission occurs unless the frequency of the incident radiation is sufficiently high.

2- When the intensity of the incident radiation increases the maximum kinetic energy with which the electrons are emitted does NOT increase.

Thanks to anyone who can shed light... I'm also now very curious as to actually why, physics homework actually leading to curiosity! wow!

You should all pay more attention in class!

As for the Qs, no idea mate!
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You should all pay more attention in class!

As for the Qs, no idea mate!

While that is certainly true, these are printed notes, not ones I've taken.

just to avoid anyone who reckons I'm just being a lazy sod. And I don't remember any mention of this in class.
Those two questions can be answered directly via the Photoelectric Effect, for which Einstein won the Nobel Prize. Instead of me giving you a semi-decent explanation, I will lead you to a more respectable source:

http://en.wikipedia.org/wiki/Photoelectric_effect
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I don't know anything about physics, but at higher education, it really helps to read around your subject a bit. Really helps you get into the mindset for uni.

I'm not saying don't ask for help, just trying to give some friendly advice that I wish I was given when I went into sixth form
1) Energy is proportional to frequency because E=hf so if the f of the incoming photon is not high enough the energy of the photon is not high enough to release an electron from the surface of the material

2)Higher intensity just means there are more photons, so more electrons released, not higher energy.

I got a B on PHY4 (this subject) in January

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1- For any surface no photoemission occurs unless the frequency of the incident radiation is sufficiently high.

Photons being massless, their energy is proportional to their associated frequency. Since photoemission depends on the photon having a certain energy there is a threshold frequency below which it doesn't happen.

2- When the intensity of the incident radiation increases the maximum kinetic energy with which the electrons are emitted does NOT increase.

At the same frequency each photon has the same energy, and for the purposes of this each electron only interacts with one photon. Increasing intensity just means there are more photos, so this means they can interact with more electrons, but the maximum amount of energy a photon can give to an electron won't increase.
He explained it all in class today, I was sitting next to you. Get yersel' sorted.

I'll be skeefing the answers anyway.

Help appreciated

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Last edited by BrokenLaney_DBC at Mar 26, 2008,
1- For any surface no photoemission occurs unless the frequency of the incident radiation is sufficiently high.
This is a "quantum" effect. Anything less than 1 equals zero. It fits the particle model of light rather well, don't you think?

2- When the intensity of the incident radiation increases the maximum kinetic energy with which the electrons are emitted does NOT increase.
Again, quantum. For EACH photon emitted, it has a specific kinetic energy. When you increase the incident radiation, you increase the probability of photons being emitted. Therefore: greater numbers of photon and greater total energy being emitted, but not greater energy for each photon.

Clear as mud, right?
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I don't quite remember, but I think electrons are emited when certain threshold of light frecuency is surpased in a surface (metals I think)....

And for 2), the emition of electrons (or number) are a principal cause of the radiation's frecuency, not intensity. I think energy isn't confined by intensity, just because the formule of photon energy is E=hc/frecuency (lambda) I think....so intensity is not necessary

Pleas correct me if I am wrong...
Last edited by gonzaw at Mar 26, 2008,
This will be hard to explain, because my physics lessons are in dutch but I will give it a shot.

1. Photoemission will only occur when the energy of the incident radiation equals or is greater than the energy needed to make an electron leave the material.

So in order for this to happen the photon will need enough energy. The formula for the amount of energy a photon has is E=h*f where h= Planck's constant and f is the frequency. As you can see, the energy of the photon will increase if f increases.

2 Each photon with a frequency high enough will release an electron. The energy of the electron only depends on the above formula, not on anything else. Increasing the intensity of the light only means it contains more photons. So it will only release more electrons but not with more kinetic energy.

I hope this will get you going
for number 2: the speed of light is constant..meaning that if there was a spaceship in outer space that was not moving and shot a light toward the earth, if you could gauge the speed of light, it would hold the value of c. Now if the spaceship was traveling at 1000 mph away from earth shot a light toward the earth again, the speedwould still hold the value of c, not 1000 mph less..so tie that into how the speed of the electrons does not increase maybe?
hmmm, interesting.

For part 1, I'd say that the incident radiation has to have a high frequency, as this will lead to high energy, so the surface will actually absord most of this energy until it reaches it's maximum, at which stage it will emit the surplus (think of a poker in the fire absorbing heat before it starts to glow).

As for part 2, if you think a Bohr model of the atom, that there are only specific values which absorbed or emitted photons can have, that shows that no matter how much energy you throw at an object, it will only emit (and, indeed, absorb) photons of specific wavelenghts, and hence energies. So although it may emit more photons, they will all have to be photons of these specific energy
I think that's sorted in the ole noggin. Hopefully. And brokenlaney, get away. away and think the leftover energy is light.

I'm not 100% sure how all these things prove the particle (or wave particle duality) nature of light... but I think my answers skirt suitably around this issue. Ta anyway.
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