# Last edited by a moderator: May 4, 2017

• stickplot
In summary: Yes, n=1 would not absorb. The energy of the photon would not be equal to the energy of the transition.
stickplot

## Homework Statement

A photon with λ = 600 nm interacts with a hydrogen atom in fundamental state. Will the photon be absorbed? Justify your answer.

## Homework Equations

hc/lambda= energy of photon
1/lambda= R(1/2^2-1/n^2) (visible light)

c= speed of light
lambda= wavelength
h= planks constant
r= rhydbergs constant

## The Attempt at a Solution

hc= 1240 eV nm
lambda=600 nm
1240/600= 2.1 Ev of photon

i don't know what to do after i get the energy of one photon and how to see if its going to get absorbed, what do i do after this? does 1/lambda= R(1/2^2-1/n^2) need to be used? and if it is what do i do after i find n?

hi,

In general the Rydberg formula for the wavelength corresponding to a transition between two levels n1 and n2 (n1>n2) is 1/lambda=R(n1-2-n2-2).

Your lambda is given, and if "fundamental" state means the ground state (n1=1) than you can solve for n2 and check whether an integer comes out. If it happens to be the case, the photon will be absorbed. Absorption occurs generally if the energy difference between the levels is equal to the energy of the photon.

Of course, in reality, things are as usual more complicated than the simple Bohr-model, but I guess that none of the complications is included in your problem

cheers

o ok. so..

1/6x10^-7=1.097x10^7(1/2^2-1/n^2)
.1519295047=1-1/n^2
1.179147259=n^2
1.08=n
so since n=1 that means that it will absorb because it is the same energy as the fundamental state?

btw.
what if it was larger like n=2? would that mean it emits part of the photon into kinetic energy?

what is important, is that is the energy difference between two levels is precisely equal to the photon energy, otherwise no absorption will occur (or to say it in a fancy way, only a kind of "virtual transition" will occur). Photons only do a "all or nothing" absorption process.

btw, it is still unclear to me, why your formula is "1/lambda= R(1/2^2-1/n^2) " ? This would suggest, that the electron is already in the n=2 state, which is not the ground state ?

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o so is it supposed to be R(1/1^2-1/n^2)??
i put R(1/2^2-1/n^2) because i assumed it was in the visible light (balmer series) because 600 nm is visible light.

wait. nm i did it the other way anyways
R(1/1^1-1/n^2) i did the formula like this up there to get n=1

stickplot said:
what if it was larger like n=2? would that mean it emits part of the photon into kinetic energy?

If you ionize the atom, then a part of the photon energy (which is not used up for ionization), will be converted into the kinetic energy of the photon. In that way, the photon will be absorbed. But if you have n=100, and the energy of the photon EPH is not sufficient to ionize the atom and EPH is not equal to the energy difference of two atomic levels, no absorption will occur.

stickplot said:
i put R(1/2^2-1/n^2) because i assumed it was in the visible light (balmer series) because 600 nm is visible light.
To evaluate this formula, you need to look at the electron levels, not at the wavelength of light. You can have absorption in visible, also if the electron levels have very high binding energy. It is only the difference between two levels that counts

stickplot said:
1.08=n
I haven't checked your calculation but as I see it, n=1.08 is not an integer ;-). If the atom would be in rest, photons with 600nm wavelength would not be absorbed. Only if you have randomly flying atoms, and 600nm is close to an energetically allowed transition, the Doppler effect might enable an absorption. But this is again a rather complicated situation.

Cheers

ok so if its off by .08 to being a integer it won't absorb??
and you said that energy of photon has to be equal to transition energy to absorb...
but isn't it always equal??
because photon= hc/lambda, but energy transition= the same thing.
so doesn't this say that it ALWAYS absorbs because it will always be equal?
thats what I am confused about :/

Transitions in one atom produce photons with specific wavelength, but these photons may not be compatible with the energy differences which occur for transition between levels in another atom of a different element. Thus in general you will have no absorption between two atoms, except by coincidence.

What you usually do in basic spectroscopy, you illuminate with a wide range of photon wavelengths. Now, the atom at which you shine the light, can only absorb specific wavelengths which correspond to its energetic transitions. If you then look at the light through a spectrometer, you will see some "absorption lines".

o ok. so this wouldn't absorb, because if n is equal to 1 it still wouldn't absorb because the energy of transition would not be equal to energy of the photon right?
hf= 2.1
energy of transition (if n=1) i= 0
and a number like 1.08 could not be used only integers right?
so that is why is no absorption occurs correct?

1. if you do not get a integer number out of the calculation , no absoprtion occurs
2. it would be advisable to use some more precise numbers for constants in your calculations, e.g. go to http://physics.nist.gov/cuu/Constants/" and check whether you are closer to an integer than with the number you have provided
3. general remark: if you get 1.08 instead of expected 1,it is your decision how you answer your question. In a simple model no absorption occurs, but you can always use more advanced concepts to argument that absoprtion occurs ;-)
4. you should refer some literature to find the limitations of the simple model we were talking about, maybe there you will find hints whether n=1.08 is still acceptable for absorption. Nothing can replace a good book ;-)

cheers and good luck

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## 1. What is a photon?

A photon is a particle of light that carries energy and has no mass. It is the smallest unit of light and is a fundamental particle in the field of quantum physics.

## 2. What is the relationship between photons and quantum physics?

Quantum physics is the branch of physics that studies the behavior and interactions of particles on a subatomic level, including photons. Photons are used to explain the phenomena of light and electromagnetism, and they play a crucial role in the quantum theory of light.

## 3. How are photons produced and detected?

Photons are produced when an atom or molecule undergoes a change in energy levels, such as during an electron transition. They can also be produced through nuclear reactions or particle interactions. Photons are detected by specialized instruments, such as photomultiplier tubes or photodiodes, which convert them into electrical signals.

## 4. What is the wave-particle duality of photons?

The wave-particle duality is a concept in quantum physics that states that particles, including photons, can exhibit both wave-like and particle-like behaviors. This means that photons can behave as waves, with properties such as frequency and wavelength, as well as particles, with properties such as energy and momentum.

## 5. How are photons used in practical applications?

Photons have a wide range of practical applications, including in medicine, telecommunications, energy production, and imaging technologies. They are used in medical imaging tools such as X-rays and MRI machines, as well as in solar panels, fiber optic communication, and laser technology.

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