Photon energy of a certain material?

In summary, a material kept at very low temperature absorbs photons with energies of 0.23 eV and 0.85 eV, but only emits photons with energies of 0.37 eV and 0.64 eV. When the material is warmed up enough to emit photons with energy of 0.64 eV, what additional photon energy is emitted is still unknown and requires further analysis and assumptions about the arrangement of energy levels in the material.
  • #1
physicslove22
27
0

Homework Statement


A certain material is kept at very low temperature. It is observed that when photons with energies between 0.23 eV and 0.85 eV strike the material, only photons of 0.37 eV and 0.64 eV are absorbed. Next the material is warmed up so that it starts to emit photons. When it has been warmed up enough that 0.64 eV photons begin to be emitted along with 0.37 eV photons, what additional photon energy is observed to be emitted by the material?

Homework Equations


-13.6/N^2 = Energy in eV

The Attempt at a Solution


I'm completely stuck :(
 
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  • #2
Unfortunately, "I'm completely stuck" doesn't qualify as an attempt at a solution. You have to try something and show us what you tried.
 
  • #3
Would I be on the right track if I set -13.6/N^2 equal to .37, and then solve for N? Then I could use trial and error to find a multiple of N that is between .23 and .85?
 
  • #4
N= 6.06, so never mind on the previous post...
 
  • #5
So I know that photons can only be absorbed at the energy amounts from where they are to N=1, so is this problem saying that N=2 is .37 and N=3 is .64?
 
  • #6
physicslove22 said:
N= 6.06, so never mind on the previous post...
What is N? And what is 13.6?
 
  • #7
"Energy in eV" in post 1 is the energy of the states, not the energy of photons that get emitted or absorbed.
 
  • #8
Oh... Also I just realized that -13.6/N^2 is only for hydrogen atoms, I will need to devise a different approach...
 
  • #9
physicslove22 said:
So I know that photons can only be absorbed at the energy amounts from where they are to N=1, so is this problem saying that N=2 is .37 and N=3 is .64?
A photon can be absorbed by a system (atom or molecule if the difference between the energy levels of the system is equal to the energy of the photon.
When the material is at very law temperature, it is at the deepest energy level. You can call it level 1 . The photon can rise it to the next level, or to the third one, and so on, according to the photon energy. If you count the energy of the levels from the first level, you can say that the energy of the second level is 0.37 eV and that of the third level is 0.64 eV.

What about photon emission at high temperatures?
 
  • #10
Would it emit photons from N=3 to N=2, N=2 to N=1, and N=3 to N=1? So it would be .64 and .37eV?
 
  • #11
ehild said:
f you count the energy of the levels from the first level, you can say that the energy of the second level is 0.37 eV and that of the third level is 0.64 eV.

No, this won't work, because that would make the difference between the second and third levels 0.64 - 0.37 = 0.27 eV, so there would have been absorption observed of 0.27 eV photons; but the conditions of the problem exclude that. So the 0.64 eV must be the difference between the second and the third levels. (Or it could be the first and second, and 0.37 eV could be the difference between the second and third; that is actually more likely, given the usual distribution of energy levels in atoms, but we don't actually need to know which is right to solve the problem in the OP.)
 
Last edited:
  • #12
physicslove22 said:
Would it emit photons from N=3 to N=2, N=2 to N=1, and N=3 to N=1?

Yes.

physicslove22 said:
So it would be .64 and .37eV?

Those are two of the three transitions you listed. (Key question: which two?) The key is to figure out the third.
 
  • #13
We don't have to know which number the different states have.

PeterDonis said:
No, this won't work, because that would make the difference between the second and third levels 0.64 - 0.37 = 0.27 eV, so there would have been absorption observed of 0.27 eV photons; but the conditions of the problem exclude that.
Not if the second level was empty at zero temperature.I think we have to make the additional assumption that there are no forbidden transitions, otherwise we can construct systems we cannot solve.
 
  • #14
mfb said:
Not if the second level was empty at zero temperature.

Even if it starts out empty, it won't be empty once some 0.37 eV photons have been absorbed.

mfb said:
I think we have to make the additional assumption that there are no forbidden transitions

It's not a matter of forbidden transitions; it's a matter of which arrangement of energy levels is consistent with the entire statement of the problem. Given only the two photon energies absorbed, there are two possible arrangements; only the extra information in the problem statement (about the range of photon energies that were used to irradiate the object) allows you to rule out one of those two possibilities and get a unique solution.
 
  • #15
PeterDonis said:
Even if it starts out empty, it won't be empty once some 0.37 eV photons have been absorbed.
I don't think we are supposed to take this second-order effect into account.

PeterDonis said:
It's not a matter of forbidden transitions; it's a matter of which arrangement of energy levels is consistent with the entire statement of the problem.
Forbidden transitions increase the number of possible arrangements of energy levels.

PeterDonis said:
Given only the two photon energies absorbed, there are two possible arrangements; only the extra information in the problem statement (about the range of photon energies that were used to irradiate the object) allows you to rule out one of those two possibilities and get a unique solution.
I don't see how the two possible arrangements lead to different predictions, with and without taking second-order effects into account. I get two different arrangements but the same answer for both.
 
  • #16
mfb said:
I don't think we are supposed to take this second-order effect into account.

If we don't, I don't see how we can solve the problem. (Feel free to PM me to discuss further, since we should be giving the OP a chance to find the solution for himself.)
 
  • #17
PeterDonis said:
No, this won't work, because that would make the difference between the second and third levels 0.64 - 0.37 = 0.27 eV, so there would have been absorption observed of 0.27 eV photons; but the conditions of the problem exclude that. So the 0.64 eV must be the difference between the second and the third levels. (Or it could be the first and second, and 0.37 eV could be the difference between the second and third; that is actually more likely, given the usual distribution of energy levels in atoms, but we don't actually need to know which is right to solve the problem in the OP.)

The only possibility I can see from this is that one of them would be 1.11, which doesn't fit the limits the problem gave us...
 
  • #18
physicslove22 said:
The only possibility I can see from this is that one of them would be 1.11

I think you mean 1.01, i.e., 0.64 + 0.37, correct?

physicslove22 said:
which doesn't fit the limits the problem gave us...

The limits in the problem were on the range of photon energies used to test for absorptions of photons. There is no reason why emissions of photons could only take place within that range of energies. The energies of photons used to irradiate the material are under the experimenter's control, so the experimenter can decide to only test for absorption of photons in some particular range of energies. But the energies of photons emitted by the material when it is heated are not under the experimenter's control; there is no way for the experimenter to restrict emissions to a particular range of energies.
 
  • #19
Oh! That makes a lot more sense now! Thanks for all your help!
 
  • #20
physicslove22 said:

Homework Statement


A certain material is kept at very low temperature. It is observed that when photons with energies between 0.23 eV and 0.85 eV strike the material, only photons of 0.37 eV and 0.64 eV are absorbed. Next the material is warmed up so that it starts to emit photons. When it has been warmed up enough that 0.64 eV photons begin to be emitted along with 0.37 eV photons, what additional photon energy is observed to be emitted by the material?
:(
Very low temperature means that only the ground level is occupied, all atoms/molecule are on that level. The absorbed photon can excite the system onto the second or third level. Usually the intensity of the light is not enough to excite many atoms/molecules, and the lifetime of an excited state is not long, so you can take that almost all particles of the material are in the ground state at low temperature. So there probability that the photon rises the system from the second level to the third one is very low.

At high enough temperatures the thermal energy of the particles is enough to excite an other particle from the first level to the second one. The number of the excited particles becomes so much that you observe the radiation when the excited particles return to their ground state. Warming further, the number of particles excited to the third level becomes so high that you observe the radiation accompanying transition from the third level to the first one. At high temperatures, there is an appreciate amount of particles excited both to the second and the third level. If the transition 3-->2 is allowed, you will observe it, and the photon emitted during this transition has the difference of the energies.

Of course, there is some probability that particles are excited to higher levels at high temperature, and then photons are emitted when transition occurs between any two levels. But the text says that the material is warmed up to the temperature, when 0.64 eV photons begin to be emitted. It means that the ratio of particles excited to higher levels is very low, so the number of photons produced from transitions involving higher levels are too few to be observed.
 

1. What is photon energy?

Photon energy refers to the energy carried by a single photon, which is a unit of light or electromagnetic radiation. It is a measure of the amount of energy contained in a particular wavelength or frequency of light.

2. How is photon energy related to the material it passes through?

The photon energy of a material depends on its composition and physical properties. Different materials have different energy levels, and thus, they interact with photons in different ways.

3. How is photon energy measured?

Photon energy is typically measured in units of electron volts (eV) or joules (J). These units indicate the amount of energy that a single photon carries.

4. Can the photon energy of a material be changed?

The photon energy of a material is determined by its physical properties and cannot be changed. However, the material's interaction with photons can be altered by external factors, such as temperature or pressure.

5. How does the photon energy of a material affect its properties?

The photon energy of a material can impact its properties, such as its color, transparency, and ability to absorb or reflect light. Materials with higher photon energy may have different properties than those with lower photon energy.

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