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Best_Mathematician
Photons help
use plancks equation: E=hc/lambda
Well use plank's equations.....however answer depends on geometry of molecule in case of part B & part C
They have given you the energy of \(E_2\), so by using equation \(E=hf\) and \(c=f \lambda\), we can form \[E_2=\frac{hc}{\lambda} \\ \\ \lambda=\frac{hc}{E_2}\] h=plank's constant c=speed of light \(E_2\)=energy Use that to find the wavelength at that energy
so is it 6.62*10^-34/3.2? hey what is c then?
"c" is the speed of light, which has a value of \(3 \times 10 ^8m/s\) and keep in mind that, E is not just 3.2, because its eV, you have to multiply by the charge too, \[\lambda=\frac{6.63 \times 10^{-3}(3 \times 10^8)}{3.2e}\]
the charge, e, is \(1.6 \times 10^{-19}C\)
so is the answer 1.24*10^-6
How do you get that? should be \[\lambda=\frac{6.63 \times 10^{-34}(3 \times 10^8)}{3.2(1.6 \times 10^{-19})}\]
ya and then i solved it using calculator...i got\[3.89 * 10^{-7}\]
and the unit of wavelength is Hz right
Hertz is for frequency, Wavelength is meter,
3.89∗10^(−7) meters ???
yes if you want to convert that into nanometers, then its \[(3.89 \times 10^{-7}) \times 10^9 = 389nm\]
oh ya thx...what abt 2nd one
The second one is referring to work function energy \(\phi\), work function is the minimum energy needed to remove an electron from a solid to a point immediately outside the solid surface. We express that by \[\phi=hf_o\] \(f_o\)= minimum frequency to cause a photoelectric emission
So the frequency plays a role here on the emission of photons.
I gotta go now, I'll leave it here, think about that :)
this is beyond my learning curve :) i can recall levers and rotations and such
haha thx tho
I would need to refresh my memory in modern physics. So no. Lol
well looks like audience is here...but the players are not. haha
I will make some researches and see if i can help :3
can't you use kirkchoffs laws here?
A couple of things... first, if you just plug in the energy E2, you're going to get the wrong answer. You instead need to use the energy difference E2-E0.
i really dont know how to do this...so if anyone can ex[lain it will be great
i do think we r using planck's equation and stuff here
Okay here's way it works. This molecule has three energy levels that you're considering -- E0, E1, E2. These are the only energies it can have -- nothing in between. Photons are just traveling packets of energy. Therefore, if the atom is in the ground state E0, it might absorb a passing photon and be promoted up to a higher energy level. This will ONLY work, though, if the photon is carrying exactly as much energy as is required. If the photon has too much energy, it won't work. If it has too little, it won't work either. The amount of energy required to promote the atom from state E0 to state E2 is E2-E0.
For a photon, \[ E_{photon} = \frac{hc}{\lambda} \] or \[ \lambda = \frac{hc}{E_{photon}} \] This tells you that if you want to promote the atom from state E0 to state E2, you need a photon of wavelength \[ \lambda = \frac{hc}{E_2-E_0} \]
is this first answer?
so the guy before who explained me the answer is wrong? we would need to include E2-E0 to get wavelength
Yes. His equations weren't wrong but you need the energy difference, not just the energy of the excited state.
For the second part, because the only allowed energies are E0, E1, and E2, the only possible promotions are from E0->E1, E0->E2, and E1->E2. Each transition has a particular energy difference. Therefore, each transition can be caused only by a photon of a precise energy. It is these photons that can be absorbed by your atom. Do you understand?
so energy photons absorbed are all of them?
Photons could only be absorbed if that have E = (E2-E0), E = (E1-E0), or E= (E2-E1).
ohk...so thats the second answer...what abt third one
The third is the reverse process. If the atom is in an excited state, it can only transition down in precise ways. That is, the only allowed downward transitions are E2->E1, E2->E0, and E1->E0. When an atom undergoes a downward transition, it emits a photon whose energy is equal to the energy difference of the transition.
oh so in last two answers we dont need any calculations and stuff?
what abt work function...is it the same thing as absorbing and emitting photons
No. The work function is not related to your question at all. It is a separate effect.
nice thanks...
wish to give u 3 medals but just have 1
Quite alright, medals are not important.
exactly thx for explaining stuff