Solving Photon Energy with Equations: Am I Right?

[meher4real]

Homework Statement

Determine the heat delivered to the interaction space in three minutes, if the amount of 5.6.10^18 photons per 1 s hits the area of 1 cm2 and the absorption coefficient for a given wavelength is 0.32.

Relevant Equations

E=hc/wavelength
N(interaction)=N0(1-e^-µx) and E(ab)= N(interaction) x µ(ab) x E

Hi Guys !
Previous question was to determine the energy of photon with wavelength = 820nm
I found that E = 2.42x10^-19 J = 1.51 ev using the equation E=hc/wavelength
- I tried to solve the above problem with various equations but all failed.
The closest equation i tried is N(interaction)=N0(1-e^-µx) and E(ab)= N(interaction) x µ(ab) x E
Please i want to know if I'm on the right equation to solve the problem ?!

 

[kuruman]

Although μ in the context you present is called the absorption coefficient, it doesn't look like you want to go there because no distance x is given through which photons move. Also, the number 0.32 has no units. I suspect it is an absorption efficiency, i.e. 32 % of the incident photons are absorbed and converted into heat.

 

[meher4real]

Thank you for your response Kuru !
Exactly, x unknown here and i was thinking that 0.32 was in meter.
Is there any other equation to solve it ? Because the next question asking if the heat is enough to ignite a paper of 0.1mm thick. i don't see the interconnection between the questions in the problem !

 

[kuruman]

I gave you my interpretation on the basis of the statement of the problem that you posted. Is this statement exactly as was given to you or is the original statement different, e.g. from another language that you translated?

 

[meher4real]

- The problem is assembly of three questions (it's exactly as was given in english)
the first was to calculate the energy of the photon the second is the above and the third is (Is the supplied heat enough to ignite a sheet of classic office paper 0.1 mm thick and with a (specific surface area) weight of 80 g.m-2? (cp = 1.34 kJ.kg-1.K-1))

 

[kuruman]

Perhaps the answer to the second part is intended to be a mathematical expression in the form $$N= N_0\left (1-e^{-\mu~x}\right)$$where $N$ is the number of photons absorbed to a depth $x$ from the surface. You have a length of 1 mm in part 3 so you can calculate the number of photons absorbed. The question remains whether 0.32 is in m^-1, cm^-1 or mm^-1. Perhaps you can do a web search and find out what's reasonable and what isn't. While you're at it, look up the combustion temperature for paper. You will need it for part 3.

 

[meher4real]

Perhaps the answer to the second part is intended to be a mathematical expression in the form $$N= N_0\left (1-e^{-\mu~x}\right)$$where $N$ is the number of photons absorbed to a depth $x$ from the surface. You have a length of 1 mm in part 3 so you can calculate the number of photons absorbed. The question remains whether 0.32 is in m^-1, cm^-1 or mm^-1. Perhaps you can do a web search and find out what's reasonable and what isn't. While you're at it, look up the combustion temperature for paper. You will need it for part 3.

That's exactly why i asked for help here, bacause I've tried using that equation but the result is to big
I took 0.32 m^-1 as absorption coeff, 0.1x10^3 m as length, 5.6x10^18 x 180s as N0
i get from that 1.01x10^21 ?
Then I'm thinking on this equation to determine the E(absorption) = N(interactions) x µ x E(photon) ?
Any thoughts ?

 

[kuruman]

1 mm =1×10^-3m not 10³m (which is 1 kilometer). Would that make a difference?

 

[meher4real]

I think it's still big value in my opinion (i calculate it for 3mins not for 1s)
3.23x10^17

 

[kuruman]

I think it's still big value in my opinion (i calculate it for 3mins not for 1s)
3.23x10^17

What does the number 3.23x10^17 represent? There are no units.

 

[meher4real]

What does the number 3.23x10^17 represent? There are no units.

Sorry for the delay, i was working.
I think it's "ev"
Did my result looks legit ? and what equation do i need to use to dermine the enough heat using weight ? I'm lost here !

 

[kuruman]

OK, it's eV. What kind of energy is it? What equation did you use to get it?

To find if there is enough heat, you need to find the temperature change of the paper if all the energy of the photons that are absorbed goes into raising the temperature of the paper. What equation do you know that relates heat added to temperature change?

 

[meher4real]

OK, it's eV. What kind of energy is it? What equation did you use to get it?

To find if there is enough heat, you need to find the temperature change of the paper if all the energy of the photons that are absorbed goes into raising the temperature of the paper. What equation do you know that relates heat added to temperature change?

As i mentioned before i used E=hc/wavelength to determine the Energy.
Can you explain more ? What do you mean by "To find if there is enough heat, you need to find the temperature change of the paper if all the energy of the photons that are absorbed goes into raising the temperature of the paper."

 

[kuruman]

As i mentioned before i used E=hc/wavelength to determine the Energy.

When you use E = hc/λ, that is the energy of a single photon. Are you saying to me that 3.23x10^17 eV is the energy of a single photon? I think not. You already found that the energy of a single photon is 1.51 eV and I agree with that number. So what energy does 3.23x10^17 eV represent?

Can you explain more ? What do you mean by "To find if there is enough heat, you need to find the temperature change of the paper if all the energy of the photons that are absorbed goes into raising the temperature of the paper."

This suggestion is for part 3. What strategy will you follow to find if "the supplied heat (is) enough to ignite a sheet of classic office paper 0.1 mm thick and with a (specific surface area) weight of 80 g.m-2? (cp = 1.34 kJ.kg-1.K-1))"

 

[meher4real]

When you use E = hc/λ, that is the energy of a single photon. Are you saying to me that 3.23x10^17 eV is the energy of a single photon? I think not. You already found that the energy of a single photon is 1.51 eV and I agree with that number. So what energy does 3.23x10^17 eV represent?

This suggestion is for part 3. What strategy will you follow to find if "the supplied heat (is) enough to ignite a sheet of classic office paper 0.1 mm thick and with a (specific surface area) weight of 80 g.m-2? (cp = 1.34 kJ.kg-1.K-1))"

I Think that 3.23x10^17 eV is the transmitted energy.

 

[kuruman]

How did you calculate that number? Show me the equation and the numbers you put in it.

 

[meher4real]

N(interaction)=N0(1-e^-µx)
where 0.32 m^-1 as absorption coeff, 0.1x10^-3 m as length (paper 0.1 mm thick), 5.6x10^18 x 180s as N0
Correction : I found 1.03x10^13 ev

 

[kuruman]

I Think that 3.23x10^17 eV is the transmitted energy.

Is that what you want? How are you going to use the transmitted energy to find if there is enough energy to set the piece of paper on fire?

 

[meher4real]

Is that what you want? How are you going to use the transmitted energy to find if there is enough energy to set the piece of paper on fire ?

The only think I'm sure about at this point that i have to use (cp = 1.34 kJ.kg-1.K-1)) and weight on my next equation.
But don't know how because I'm thinking about : E(absorption) = N(interactions) x µ x E(photon)

 

[kuruman]

The question in part 3 is:
Is the supplied heat enough to ignite a sheet of classic office paper 0.1 mm thick and with a (specific surface area) weight of 80 g.m-2? (cp = 1.34 kJ.kg-1.K-1))?

What strategy is needed to answer this question? By this I mean what do you need to know? Is calculating the absorbed energy sufficient? If not, what else do you need to know or look up or calculate? Hint: Not all things that absorb energy ignite. Why not?

 

[meher4real]

The question in part 3 is:
Is the supplied heat enough to ignite a sheet of classic office paper 0.1 mm thick and with a (specific surface area) weight of 80 g.m-2? (cp = 1.34 kJ.kg-1.K-1))?

What strategy is needed to answer this question? By this I mean what do you need to know? Is calculating the absorbed energy sufficient? If not, what else do you need to know or look up or calculate? Hint: Not all things that absorb energy ignite. Why not?

Sorry for the delay i was at work !
Thank you for your time Kuru, but as i said before I'm a little confused because i don't know an equation that needs CP? plus i find some problems understanding some scientific sentences, I'm trying my best by using different equations to solve it but still no results.
Only one problem left for my exam and this took me too much time.
I need your help with more details, I'm not a physician by the way and i used to study physics in french.

 

[kuruman]

I am not asking to find an equation. I am asking for a plan. Once you have a plan, you will know where to look for equations. If this is an exam problem, the solution must come from you. Here is a simple question, no equation is needed. It's the first step towards understanding what comes next.
Question: As you know, a piece of paper does not ignite if it is just sitting there minding its own business. What must happen to it in order to ignite?

 

[meher4real]

I am not asking to find an equation. I am asking for a plan. Once you have a plan, you will know where to look for equations. If this is an exam problem, the solution must come from you. Here is a simple question, no equation is needed. It's the first step towards understanding what comes next.
Question: As you know, a piece of paper does not ignite if it is just sitting there minding its own business. What must happen to it in order to ignite?

Understood !
Must expose it to heat, means must absorb energy ?

 

[kuruman]

Yes. But simply absorbing energy is not enough to ignite something. A piece of paper that just sits in the Sun does not catch fire. If, however, you focus the sunlight on it with a magnifying glass, it will catch fire. Why the difference?

 

[meher4real]

Yes. But simply absorbing energy is not enough to ignite something. A piece of paper that just sits in the Sun does not catch fire. If, however, you focus the sunlight on it with a magnifying glass, it will catch fire. Why the difference?

T° increase

 

[kuruman]

T° increase

Good. Now you have to answer two questions.
1. How is the rise in temperature related to the absorbed energy? Hint: You are given c_p. That is known as the specific heat. Do some research on the word and you will come up with an equation.
2. The temperature of both pieces of paper increases. Why does one ignite and the other does not?

 

[meher4real]

Good. Now you have to answer two questions.
1. How is the rise in temperature related to the absorbed energy? Hint: You are given c_p. That is known as the specific heat. Do some research on the word and you will come up with an equation.
2. The temperature of both pieces of paper increases. Why does one ignite and the other does not?

Thank you Kuru !
I'll be back with the results.

 

[meher4real]

Thank you Kuru !
I'll be back with the results.

By the way, is N found is the absorption energy ? is it the right result for part 2 ?

 

[kuruman]

By the way, is N found is the absorption energy ? is it the right result for part 2 ?

N is the number of absorbed photons. How would you find the absorbed energy if you know the number of absorbed photons and the energy of one photon?

 

[meher4real]

N is the number of absorbed photons. How would you find the absorbed energy if you know the number of absorbed photons and the energy of one photon?

 

[kuruman]

I can't read that. There are too many erasures. Can you type it up, please?

 

[meher4real]

I can't read that. There are too many erasures. Can you type it up, please?

- N(inter)=5.6x10^18 x 180 x (1-e^(-0.32x0.1x10^-3)) = 1.03x10^13 photons
- E(abs) = N(inter)xE(photon)
= 2.50 x 10^-6 J
- Energy required to heat the paper :
Q=m x Cp x T = 80x10^-3kg x 1.34 J/kg x 233° = 24.97 J
- Number of photons required :
24.97/2.42x10^-19 J = 1.032x10^20 photons

 

[kuruman]

- N(inter)=5.6x10^18 x 180 x (1-e^(-0.32x0.1x10^-3)) = 1.03x10^13 photons

This number is too low by several orders of magnitude. The left-hand side is OK, the right-hand side is not. You must have made a mistake somewhere.

 

[meher4real]

This number is too low by several orders of magnitude. The left-hand side is OK, the right-hand side is not. You must have made a mistake somewhere.

i used the thickness of the paper which is 0.1 mm ?! you said that is 1 mm can you explain why ?

 

[kuruman]

You used the correct thickness of the paper 0.1 mm as given by the problem. What I am saying is that 5.6x10^18 x 180 x (1-e^(-0.32x0.1x10^-3)) is NOT 1.03x10^13. You need to redo this calculation correctly.