# How can i find the absorption/emission transitions from Energy levels

• hbk69
In summary: Just one last query, when i am calculating ΔE for absorption transitions do i always subtract the Final energy from the Initial energy i.e. ΔE = Efinal of atom - Einitial of atom and for emission transitions do i always subtract the Initial energy from the Final energy i.e. ΔE = Efinal of atom - Einitial of atom?Delta E is a number that represents a difference between two energies. The difference can be positive or negative, depending on the specific energies. The formula ΔE = Efinal - Einitial applies to both absorption and emission. You just have to make sure that you are using the correct energies for each situation.Delta E is a number that represents a difference between
hbk69

## Homework Statement

Question 1)

For absorption transitions:

At n=1 the energy is 0.0eV (electron volts), at n=2 the energy is 3eV, at n=3=5eV

Question 2)

For emission transisions:

At n=1 the energy is 0.0eV, n=2 energy is 3eV and at n=3 energy is 5eV

Question 3)

The atom has energy levels n=1(0eV),n=2(2eV) and n=3(5eV). A photon has energy E=3eV which corresponds to spectral line with λ=414nm. Where would this spectral line be observed? absorption spectrum, emission spectrum, both or neither?

## Homework Equations

ΔE between the energy levels for an absorption transition ΔE= E1 - E2 etc..
ΔE between the energy levels for an emission transition ΔE=E3 - E2 etc..

λ=c*h/ΔE

## The Attempt at a Solution

Question 1)

So to find the difference in energy ΔE between the energy levels for an absorption transition at ΔE=E1 - E2, then ΔE= 0.0eV - 3eV = -3eV but in my notes this answer is positive 3eV what am i doing wrong?

Question 2)

To find emission transition at ΔE=E3 - E2 , then ΔE= 5eV-3eV= 2eV but in my lecture notes the answer given is negative -2eV i do not see how this is possible and where i went wrong?

Question 3)

Since E=3eV, I would assume it would be an emission spectrum because ΔE=E3 - E2=3eV but shouldn't the 3eV be negative because for an emission transion ΔE=E3 - E2 in question 2 the answer given was negative.

You sound like the guy in the joke that was holding a pencil with the lead end up an wondering why the lead was at the wrong end of the pencil. Try flipping the pencil around.

If an atom absorbs a photon, the energy of the photon is transferred to the atom, therefore one has ##\Delta E >0## for the atom. Conversely, when an atom emits a photon, the photon carries away some of the energy of the atom, and therefore ##\Delta E < 0##.

If the emitted photon has an energy of +3 eV, by how much did the energy of the atom change?

DrClaude said:
If an atom absorbs a photon, the energy of the photon is transferred to the atom, therefore one has ##\Delta E >0## for the atom. Conversely, when an atom emits a photon, the photon carries away some of the energy of the atom, and therefore ##\Delta E < 0##.

If the emitted photon has an energy of +3 eV, by how much did the energy of the atom change?

I do not understand your question properly but am guessing -3eV since you said for emissions transitions are ##\Delta E < 0## does that mean for absorption transition its always E2 -E1 so i get a positive answer and E1 - E2 so i get a negative answer for emission transition for questions 1 and 2. But then in question 3 i need to find the type of transition for E=3eV since this is positive it would an absorption transition. That can not be possible since absorption transition always begin from ground state. Confused lol

How can i solve the issues in the OP with the three questions please any help appreciated. Thanks

The symbol ΔE represents the change in energy of the atom.

So, ΔE = Efinal of atom - Einitial of atom.

This is true for both absorption and emission.

You just have to make sure that, for a specific transition, you know which energy level is the initial level and which is the final level.

1 person
TSny said:
The symbol ΔE represents the change in energy of the atom.

So, ΔE = Efinal of atom - Einitial of atom.

This is true for both absorption and emission.

You just have to make sure that, for a specific transition, you know which energy level is the initial level and which is the final level.

Thanks for the help :), makes sense now in regards to my problems with question 1 and question 2. But in question 3:

The atom has energy levels n=1(0eV),n=2(2eV) and n=3(5eV). A photon has energy E=3eV which corresponds to spectral line with λ=414nm. Where would this spectral line be observed? absorption spectrum, emission spectrum, both or neither?

Since it is obvious that the photon of energy E=3eV belongs to an emission spectrum why is the 3eV positive? because since ΔE = Efinal of atom - Einitial of atom in this case it would be the Final energy 2eV - Initial energy 5eV = -3eV

hbk69 said:
The atom has energy levels n=1(0eV),n=2(2eV) and n=3(5eV). A photon has energy E=3eV which corresponds to spectral line with λ=414nm. Where would this spectral line be observed? absorption spectrum, emission spectrum, both or neither?

Since it is obvious that the photon of energy E=3eV belongs to an emission spectrum why is the 3eV positive? because since ΔE = Efinal of atom - Einitial of atom in this case it would be the Final energy 2eV - Initial energy 5eV = -3eV

The problem is not well formulated. Of course, if you are detecting a photon of energy 3 eV, that would correspond to emission. But a "spectral line at λ=414nm" doesn't necessarily mean that a photon of that energy has been detected.

By convention, photon energies are taken to be positive, ##E = h \nu##. The ##\Delta E## for the atom will be negative if the photon was emitted, and positive if the photon was absorbed.

1 person
DrClaude said:
The problem is not well formulated. Of course, if you are detecting a photon of energy 3 eV, that would correspond to emission. But a "spectral line at λ=414nm" doesn't necessarily mean that a photon of that energy has been detected.

By convention, photon energies are taken to be positive, ##E = h \nu##. The ##\Delta E## for the atom will be negative if the photon was emitted, and positive if the photon was absorbed.

Thanks for the help, i understand the issues i had earlier.

## 1. How do energy levels affect absorption and emission transitions?

Energy levels play a crucial role in absorption and emission transitions. When an atom or molecule absorbs energy, its electrons move to higher energy levels. Similarly, when an electron in a higher energy level drops down to a lower level, energy is released in the form of light, causing an emission transition.

## 2. How can I determine the energy levels of an atom or molecule?

The energy levels of an atom or molecule can be determined by using spectroscopy techniques such as absorption and emission spectroscopy. These methods involve passing light of different wavelengths through a sample and measuring the amount of energy absorbed or emitted. The resulting spectrum can then be used to identify the energy levels.

## 3. Can the absorption and emission transitions of an atom or molecule be predicted?

Yes, the absorption and emission transitions of an atom or molecule can be predicted using quantum mechanics. The energy levels and transitions are determined by the electronic structure of the atom or molecule, which can be calculated using quantum mechanical equations and principles.

## 4. How can I find the specific absorption and emission transitions of a particular atom or molecule?

To find the specific absorption and emission transitions of a particular atom or molecule, you can consult a database or reference book that contains the energy level diagrams and corresponding transitions. These references also often provide information on the wavelengths and energies of the transitions.

## 5. What is the relationship between the energy of a transition and the wavelength of light emitted or absorbed?

The energy of a transition is directly proportional to the wavelength of light emitted or absorbed. This relationship is described by the equation E = hc/λ, where E is the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. This means that transitions with higher energies correspond to shorter wavelengths of light, and vice versa.

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