How much energy in light is related to what?

[BadBoy]

Hi

First off I apologise that this is an easy high school question. I have actually tried to google the answer last night but didnt come up with anything (well, anything I could understand anyway).

Now, I have a degree is Electronics and did a bit on Electro Magnetic Radiation, impedance matching and transmission lines etc. However it was a long time ago and I was never that good at it anyway.

My child is at high school and lost a mark on a Science test and she asked me why. The question asked was:

Infra Red is used in a toaster and a TV remote. Why dont we burn ourselves is we point the remote at our skin.

Her answer was that infrared has a low amount of energy. and that would be my answer. But it got me thinking on how energy is related to EMR.

Looking at various resources which show the electro magnetic spectrum against energy, the energy increases with frequency.

Microwaves have a longer wavelength than infrared and so less energy.

So why do we have microwave ovens and not infrared ovens. And why is the energy given from an infrared remote control so low is doesnt burn us. How can we cook things with Microwaves if they have less energy than infrared, which doesnt burn us when coming from a TV remote.

My immediate thought was to do with amplitude, or in effect area under the curve (sine wave)? Is this correct?

 

[jj]

An IR remote has a very low radiated power, in other words, the number of photons emitted by the remote is many times smaller than the number of photons emitted by the glowing wire.

A photon of a given wavelength has a defined energy, but something can emit one, or 10^20th, which obviously will transmit a lot more energy, via a lot more photons.

 

[BadBoy]

An IR remote has a very low radiated power, in other words, the number of photons emitted by the remote is many times smaller than the number of photons emitted by the glowing wire.

A photon of a given wavelength has a defined energy, but something can emit one, or 10^20th, which obviously will transmit a lot more energy, via a lot more photons.

ok, but we have to consider this is terms of particals rather than waves?

Also, I guess the visible part of the spectrum in a toaster is what cooks the bread, rather than the infrared? I mean the questions a bit misslading is it not because the infrared bit even in a toaster isnt whats doing the damage?

 

[egslim]

Her answer was that infrared has a low amount of energy.

Infrared means the photons have low amount of energy each, if a device emits a large enough number of them you will still get burned. As with the toaster that was mentioned.

She should have said the IR remote has a low amount of energy.

ok, but we have to consider this is terms of particals rather than waves?

Not necessarily, classic electromagnetic theory is perfectly able to explain the difference. But it's much easier with photons.

 

[jj]

ok, but we have to consider this is terms of particals rather than waves?

Also, I guess the visible part of the spectrum in a toaster is what cooks the bread, rather than the infrared? I mean the questions a bit misslading is it not because the infrared bit even in a toaster isnt whats doing the damage?

Doesn't matter. NO, it's the infrared primarily, there's just a lot more photons coming from the toaster than there are coming from the remote. Each photon of a given wavelength carries a specific amount of energy, so 10 of them is 10 times the energy, and so on.

 

[BadBoy]

Doesn't matter. NO, it's the infrared primarily, there's just a lot more photons coming from the toaster than there are coming from the remote. Each photon of a given wavelength carries a specific amount of energy, so 10 of them is 10 times the energy, and so on.

I thought I read someplace that something that is glowing cherry red would be giving out more energy at that part of the spectrum than the infrared bit?

 

[BadBoy]

I thought I read someplace that something that is glowing cherry red would be giving out more energy at that part of the spectrum than the infrared bit?

Thanks for the feedback BTW. Makes sense. A remote with 2*AA batteries isnt going to be able to generate much power, I just didnt understand how it related to frequency.

So, for a microwave source and a infrared source that are able to emit the same number of photons per second, the infrared radiation will carry more energy.

 

[Brian-M]

ok, but we have to consider this is terms of particals rather than waves?

Light consists of photons, which are wavelike particles. So yes, you have to consider it in terms of particles.

Multiply the number of photons per second by the energy per photon to get the power. Infrared photons may have less energy than visible photons, but if an object is radiating more infrared photons than visible photons, you can have most of the radiant energy in the infrared range.

Also, I guess the visible part of the spectrum in a toaster is what cooks the bread, rather than the infrared? I mean the questions a bit misslading is it not because the infrared bit even in a toaster isnt whats doing the damage?

You guess wrong. Look the element while the toaster is in use, it's emitting a dim red light. Is that enough to toast with?

In order to cook toast with visible-light in a reasonable amount of time, the light would have to be so bright that just looking at it would send you blind.

I thought I read someplace that something that is glowing cherry red would be giving out more energy at that part of the spectrum than the infrared bit?

If you did, your source was wrong. Think of the frequencies being radiated by the element as being in a bell curve. When something is glowing red from heat, that means the right hand edge of the bell curve is poking into the visible region. But most of the energy is in the center of the bell, well down in the infrared region.

 

[BadBoy]

Light consists of photons, which are wavelike particles. So yes, you have to consider it in terms of particles.

Multiply the number of photons per second by the energy per photon to get the power. Infrared photons may have less energy than visible photons, but if an object is radiating more infrared photons than visible photons, you can have most of the radiant energy in the infrared range.



You guess wrong. Look the element while the toaster is in use, it's emitting a dim red light. Is that enough to toast with?

In order to cook toast with visible-light in a reasonable amount of time, the light would have to be so bright that just looking at it would send you blind.



If you did, your source was wrong. Think of the frequencies being radiated by the element as being in a bell curve. When something is glowing red from heat, that means the right hand edge of the bell curve is poking into the visible region. But most of the energy is in the center of the bell, well down in the infrared region.

ok, thanks for that. Makes much more sense now.

Thanks for your replies guys (and ladies?)

 

[The Man]

I thought I read someplace that something that is glowing cherry red would be giving out more energy at that part of the spectrum than the infrared bit?

Yes, you are (or they were) correct, in part. Though the "infrared bit" is the same.


http://en.wikipedia.org/wiki/Planck_constant#Black-body_radiation

Which is to say the energy is related to freqauncy by the Planck constant. Which was part of the problem back then (before quantum mechanics). If you look at this graph you will see that that the emitted intensity peaks at a particular wavelength (by temperature) at about 500 nm. Just under the 750 nm (general) limit for infrared.

http://en.wikipedia.org/wiki/Infared

So yes the hotter it is the more energy it emits at that (the infrared) part of the spectrum. However, energy can be emitted at other parts of the spectrum as well as from accelerated charged particles. Accretion disc X-ray emissions for example.

http://en.wikipedia.org/wiki/Accretion_disc


http://en.wikipedia.org/wiki/X-ray#Sources

http://arxiv.org/abs/0903.0978

ETA:

Oops, I see others have covered the primary portion

 

[BadBoy]

Yes, you are (or they were) correct, in part. Though the "infrared bit" is the same.


http://en.wikipedia.org/wiki/Planck_constant#Black-body_radiation

Which is to say the energy is related to freqauncy by the Planck constant. Which was part of the problem back then (before quantum mechanics). If you look at this graph you will see that that the emitted intensity peaks at a particular wavelength (by temperature) at about 500 nm. Just under the 750 nm (general) limit for infrared.

http://en.wikipedia.org/wiki/Infared

So yes the hotter it is the more energy it emits at that (the infrared) part of the spectrum. However, energy can be emitted at other parts of the spectrum as well as from accelerated charged particles. Accretion disc X-ray emissions for example.

http://en.wikipedia.org/wiki/Accretion_disc


http://en.wikipedia.org/wiki/X-ray#Sources

http://arxiv.org/abs/0903.0978

ETA:

Oops, I see others have covered the primary portion

So (and sorry for being so slow) what the graph says is that for the temperature of a toaster, the peak radiance is going to be in the infrared spectrum.

However if we kept increasing the temperature, the infrared energy would drop off and the energy given off at the visible light wavelength would increase (and eventually drop off if we got it hot enough?).

 

[The Man]

Thanks for the feedback BTW. Makes sense. A remote with 2*AA batteries isnt going to be able to generate much power, I just didnt understand how it related to frequency.

So, for a microwave source and a infrared source that are able to emit the same number of photons per second, the infrared radiation will carry more energy.

Yes, per photon infrared has more energy than microwaves, but the modes of cooking are different.

http://en.wikipedia.org/wiki/File:EM_Spectrum_Properties_edit.svg

Infrared heats from the surface inwards. While microwaves osculate the water molecules (by their dipole configuration) to heat from the inside out.

http://en.wikipedia.org/wiki/Microwave_oven#Principles

http://en.wikipedia.org/wiki/Dielectric_heating

 

[The Man]

So (and sorry for being so slow) what the graph says is that for the temperature of a toaster, the peak radiance is going to be in the infrared spectrum.

However if we kept increasing the temperature, the infrared energy would drop off and the energy given off at the visible light wavelength would increase (and eventually drop off if we got it hot enough?).

No the infrared peak increases. However, as average molecule velocities increase more energy can be emitted by shorter (such as X-ray) wavelengths.


Try Hyperphysics, on heat and thermodynamics

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Sorry I can't remember how to link a specific page.

 

[The Man]

Hi

First off I apologise that this is an easy high school question. I have actually tried to google the answer last night but didnt come up with anything (well, anything I could understand anyway).

Now, I have a degree is Electronics and did a bit on Electro Magnetic Radiation, impedance matching and transmission lines etc. However it was a long time ago and I was never that good at it anyway.

My child is at high school and lost a mark on a Science test and she asked me why. The question asked was:



Her answer was that infrared has a low amount of energy. and that would be my answer. But it got me thinking on how energy is related to EMR.

Looking at various resources which show the electro magnetic spectrum against energy, the energy increases with frequency.

Microwaves have a longer wavelength than infrared and so less energy.

So why do we have microwave ovens and not infrared ovens. And why is the energy given from an infrared remote control so low is doesnt burn us. How can we cook things with Microwaves if they have less energy than infrared, which doesnt burn us when coming from a TV remote.

My immediate thought was to do with amplitude, or in effect area under the curve (sine wave)? Is this correct?

Sorry just looked at your first post. I use to design connectors for electrical distribution and transmission. As it seems no one has mentioned it before, there are inferred ovens on the market. Heck if you got an electric oven, like me, it is an inferred oven, just a big arse toaster.

 

[BadBoy]

Sorry just looked at your first post. I use to design connectors for electrical distribution and transmission. As it seems no one has mentioned it before, there are inferred ovens on the market. Heck if you got an electric oven, like me, it is an inferred oven, just a big arse toaster.

aha, of course.

I found microwaves one of the single most amazing/interesting part of the course.

I remember messing around with microwaves and wave guides and measuring standing waves with some meter that slid up and down the wave guide. I seem to remember working out how to use Smith charts for impedance matching etc. I find software development much easier in many ways, which is what I do now, compared with electronics, especially analog.

 

[Tubbythin]

While microwaves osculate the water molecules (by their dipole configuration) to heat from the inside out.

As I understand it, this is not really true. A microwave will only heat something from the inside out if the something is dry on the outside and wet on the inside (e.g a pie). Then the photons will be absorbed primarily in the middle.
If the heated item is of uniform consistency then less photons reach the centre and so the outside will cook quicker.

 

[BadBoy]

No the infrared peak increases. However, as average molecule velocities increase more energy can be emitted by shorter (such as X-ray) wavelengths.


Try Hyperphysics, on heat and thermodynamics

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Sorry I can't remember how to link a specific page.

hu,

But it looks like the curves work their way into the shorter wavelengths (to the left) towards the visible light part of the spectrum, as temperature increases, and so infrared wavelengths are on a lower part of the intensity curve are they not?

 

[Cuddles]

Microwaves have a longer wavelength than infrared and so less energy.

So why do we have microwave ovens and not infrared ovens.

Well, as already noted we do have infrared ovens. But it's important to note that the total amount of energy is not the only thing that needs to be considered. Also important are things like how well different wavelengths are absorbed by the target, and how efficiently you can transmit it in the first place. Microwaves are nice for cooking for both of those. A microwave oven actually sets up a standing wave inside, so very little energy is lost other than that which is absorbed by the food. Check how hot your microwave gets compared to your oven. The oven wastes a lot of energy that goes into heating up the body of the oven, the air inside, and so on.

As for absorbency, just think about why we don't use x-rays to cook food. Sure, the photons have higher energy, but most of them will just go straight through rather than getting absorbed and heating it up. Microwaves are absorbed reasonably well by water, so they're a good choice if you want to cook food even though they're towards the lower end of the frequency spectrum. I think infrared is actually absorbed better by liquid water, but the point is that there are many different issues to consider.

As I understand it, this is not really true. A microwave will only heat something from the inside out if the something is dry on the outside and wet on the inside (e.g a pie). Then the photons will be absorbed primarily in the middle.
If the heated item is of uniform consistency then less photons reach the centre and so the outside will cook quicker.

This is correct, but The Man isn't completely wrong either. Even if something is wet, microwaves will still penetrate a bit. So you don't simply cook from the inside out, but you do tend to get more even heat distribution (more or less even distribution through the outer 3-4cm according to Wiki) than in a regular oven which can only heat from the outside.

But it looks like the curves work their way into the shorter wavelengths (to the left) towards the visible light part of the spectrum, as temperature increases, and so infrared wavelengths are on a lower part of the intensity curve are they not?

See this graph. Yes, infrared will be on the lower part of the curve as temperature increases. But increasing temperature doesn't just shift the curve, it also increases the total energy output so that the lower part of the curve is higher than the peak was at lower temperatures.

 

[BowlOfRed]

hu,

But it looks like the curves work their way into the shorter wavelengths (to the left) towards the visible light part of the spectrum, as temperature increases, and so infrared wavelengths are on a lower part of the intensity curve are they not?

Yes, they do. I believe the objection was to the phrase:

if we kept increasing the temperature, the infrared energy would drop off and the energy given off at the visible light wavelength would increase...

The phrase "drop off" makes it sound like there is less IR coming from the higher temperature object.

If you increase the temperature, *none* of the energy bands decrease. Instead, the higher energy levels increase by a greater proportion. If you were to normalize the curve, yes the longer wavelengths make up a lower percentage of the total output. But that doesn't mean there's actually less IR radiation.

Here are some radiation curves:
http://hyperphysics.phy-astr.gsu.edu/hbase/bbrc.html

Note that as the temperature increases, the peak moves to the left. But the curve doesn't decrease on the right, it's still increasing.

 

[BadBoy]

of course. I understand now, thanks. As the temperature increases so does the infrared energy, albeit being less than the visible light at higher temperatures. Thanks.