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

[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.

 

[dauto]

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.

 

[DrClaude]

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?

 

[hbk69]

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

 

[hbk69]

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

 

[TSny]

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

So, ΔE = E_{final of atom} - E_{initial 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.

 

[hbk69]

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

So, ΔE = E_{final of atom} - E_{initial 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 = E_{final of atom} - E_{initial of atom} in this case it would be the Final energy 2eV - Initial energy 5eV = -3eV

 

[DrClaude]

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 = E_{final of atom} - E_{initial 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.

 

[hbk69]

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.