How does an electron in a hydrogen atom utilize extra energy during excitation?

[UchihaClan13]

Okay so I just had another (silly) doubt
Consider a hydrogen atom (1 proton in the nucleus and 1 1s electron)
We know that the total energy of the hydrogen atom in its ground state is -13.6 eV
And let's say in order to excite the electron to jump to n=2 the total energy should be -13.6/4=-3.4eV meaning we need to supply the electron (or rather the nucleus-electron system) with 10.2eV so as to allow it to make the transition to an excited state
My question is
If we provide it say 11eV instead of 10.2 eV or 12 eV, how would it utilise the 0.8 eV or 1.8 eV of energy?
Will it increase its kinetic energy (since the kientic energy decreases as we go further away from the nucleus of the atom) or will it simply neglect the energy or extra energy?
I suppose it won't neglect it but even if the electron say, absorbs it how would it use it or expend it? (Weaken the repulsion, increase its overall motion or what?)
Some insight is much appreciated! !
And it's nice to be back!:)
UchihaClan13

 

[Borek]

Have you considered the idea that the atom won't be excited at all?

Why is the glass transparent in the visible light?

 

[ZapperZ]

Okay so I just had another (silly) doubt
Consider a hydrogen atom (1 proton in the nucleus and 1 1s electron)
We know that the total energy of the hydrogen atom in its ground state is -13.6 eV
And let's say in order to excite the electron to jump to n=2 the total energy should be -13.6/4=-3.4eV meaning we need to supply the electron (or rather the nucleus-electron system) with 10.2eV so as to allow it to make the transition to an excited state
My question is
If we provide it say 11eV instead of 10.2 eV or 12 eV, how would it utilise the 0.8 eV or 1.8 eV of energy?
Will it increase its kinetic energy (since the kientic energy decreases as we go further away from the nucleus of the atom) or will it simply neglect the energy or extra energy?
I suppose it won't neglect it but even if the electron say, absorbs it how would it use it or expend it? (Weaken the repulsion, increase its overall motion or what?)
Some insight is much appreciated! !
And it's nice to be back!:)
UchihaClan13

As Borek has stated, you are ignoring the possibility (in fact, the STRONG possibility) that no absorption occurs.

Look at the absorption spectra on this page (Fig. 6.3.6 and 6.3.7). There are photons with higher energies available, but they are NOT absorbed. Only those with the right energy are the ones absorbed, and these correspond to the same emission spectrum, i.e. when the atom decays from its excited state. This only makes sense if the atoms only absorb photons with the exact energies to allow for each of the transition, nothing more, nothing less.

Zz.

 

[UchihaClan13]

I did consider the possibility that the photon would pass unabsorbed
Otherwise we wouldn't have stimulated emission (as in lasers in which a more energetic atom releases a photon, equal in frequency to the incident photon which stimulates it and reaches a lower level)
So what're you're basically implying
That out of the energy provided to the electron, it absorbs only a discrete amount to attain excitation and doesn't care about the extra energy?
So if we have say enough energy to excite an electron to say n=4
The electron will directly reach n=4 (a direct jump, in a similar analogy to the incident electrons knocking out directly the core electrons allowing a L-electron to occupy the vacancy and release radiation in the form of x-rays)
So is that it?
Well my new question is
Why won't its kinetic (rotational and translational ) energy increase because of the extra energy
Or is it that the quantum of energy supplied alters its potential energy and kinetic energy both?
Thanks
UchihaClan13

 

[UchihaClan13]

The glass is transparent
Because the frequency of the photons is not the correct frequency required for the absorption by the atoms
Yes
Your reasoning is correct
However in my post
I classified extra energy than the discrete amount

 

[ZapperZ]

The glass is transparent
Because the frequency of the photons is not the correct frequency required for the absorption by the atoms
Yes
Your reasoning is correct
However in my post
I classified extra energy than the discrete amount

Please note that for a SOLID (as opposed to atoms), you cannot consider atomic transition for light transport. A graphite and a diamond are BOTH made up of carbon atoms. Yet, they have different index of refraction and very different optical properties. This clearly shows that you cannot explain these things simply by the atoms that make up the solid! There are phonons (collective vibration of the solids that are not present in isolated atoms) that also come into play, among other things!

Here's the thing: If a transition is available, and the energy matches, then that transition will occur. You simply cannot "speculate" about "rotational" or "translational" energy without showing the existence of these states. If this is a molecule, then you need to show these states due to the existence of extra degree of freedom. There ARE absorption spectra showing these, but you started off considering isolated atoms (H atom) and wanting to know the kinds of absorption in that case. There are no "rotational/translational" states here (solve the H-atom Schrodinger equation).

So either you just stick to your original context, or this thing will get confusing very quickly, especially when you still haven't understood that simplest case yet.

Zz.

 

[Kevin McHugh]

There are no quantized states for translations. There are for rotations but they are in the microwave region. Electronic transitions are quantized, and only certain transitions are allowed, corresponding to exact photon energy.

Just because E_total = E_trans + E_rot + E_vib + E_elec does not mean extra energy from photons contributes to these other energy terms. There is no extra energy.

 

[UchihaClan13]

Okay
I understood it
And yes
Hamiltonian times psi=E times psi
Does quantize the energy for an H-atom
And it shouldn't have any rotational degrees of freedom (only translational,)
So
There's no extra energy or anything of the sort
So back to my original question
If I let's say do supply a hydrogen atom with an energy of 11eV
It takes up the discrete 10.2eV to reach n=2 (the excited state)
And doesn't bother about the remaining 0.8 eV
Which is of no use to it
And can be analogically considered discarded
Is this right?

UchihaClan13

 

[UchihaClan13]

And also
An atom only has the translational degree of freedom
In the x, y and z directions
It can only translate
But not rotate (there's no other atom sharing a common axis for the atom to rotate about)
Or vibrate (it cannot vibrate all by itself)
So there's no quantization for translational motion or kinetic energy
So why won't the extra energy go to it?
Or wait
Is it because of schroedinger's equation of h psi= e psi
Which allots a certain energy to each electron (based on its wavefunction)
Wow
Q.M does defy general mechanics a lot!:)

Uchiha Clan13

 

[UchihaClan13]

But when we increase the temperature aren't we increasing the average translational kinetic energy of the atoms?

So, there is an increase in the translational energy, right?
For an electron (it isn't possible because of the electron being defined by a wavefunction and the variation integral which is calculated calculates the best values greater than or equal to its ground state
Schroedinger's wave equation does fix its energy only to increase or decrease it by certain amounts
And hence any arbitrary increase in the electron's K.E isn't possible)

UchihaClan13

 

[ZapperZ]

Which part of "the energy is not absorbed" did you not understand?

Zz.

 

[Ygggdrasil]

The atom can't gain any significant translational energy from the photon because of conservation of momentum.