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aaronq Group TitleBest ResponseYou've already chosen the best response.0
use plancks equation: E=hc/lambda
 one year ago

rohankv Group TitleBest ResponseYou've already chosen the best response.0
Well use plank's equations.....however answer depends on geometry of molecule in case of part B & part C
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
how?
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
so is it 6.62*10^34/3.2? hey what is c then?
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
"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, \[\lambda=\frac{6.63 \times 10^{3}(3 \times 10^8)}{3.2e}\]
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
the charge, e, is \(1.6 \times 10^{19}C\)
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
so is the answer 1.24*10^6
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
How do you get that? should be \[\lambda=\frac{6.63 \times 10^{34}(3 \times 10^8)}{3.2(1.6 \times 10^{19})}\]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ya and then i solved it using calculator...i got\[3.89 * 10^{7}\]
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
That's the wavelength :)
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
and the unit of wavelength is Hz right
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
Hertz is for frequency, Wavelength is meter,
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
3.89∗10^(−7) meters ???
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
yes if you want to convert that into nanometers, then its \[(3.89 \times 10^{7}) \times 10^9 = 389nm\]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
oh ya thx...what abt 2nd one
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
So the frequency plays a role here on the emission of photons.
 one year ago

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
I gotta go now, I'll leave it here, think about that :)
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
this is beyond my learning curve :) i can recall levers and rotations and such
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha thx tho
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
I would need to refresh my memory in modern physics. So no. Lol
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
haha ok.
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
@Frostbite might know.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
well looks like audience is here...but the players are not. haha
 one year ago

Eyad Group TitleBest ResponseYou've already chosen the best response.1
I will make some researches and see if i can help :3
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
can't you use kirkchoffs laws here?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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 E2E0.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
i really dont know how to do this...so if anyone can ex[lain it will be great
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
i do think we r using planck's equation and stuff here
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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 E2E0.
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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 \[ \lambda = \frac{hc}{E_2E_0} \]
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
is this first answer?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
so the guy before who explained me the answer is wrong? we would need to include E2E0 to get wavelength
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Yes. His equations weren't wrong but you need the energy difference, not just the energy of the excited state.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
oh gotcha
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
so energy photons absorbed are all of them?
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Photons could only be absorbed if that have E = (E2E0), E = (E1E0), or E= (E2E1).
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
ohk...so thats the second answer...what abt third one
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
oh so in last two answers we dont need any calculations and stuff?
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
what abt work function...is it the same thing as absorbing and emitting photons
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
No. The work function is not related to your question at all. It is a separate effect.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
nice thanks...
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
wish to give u 3 medals but just have 1
 one year ago

Jemurray3 Group TitleBest ResponseYou've already chosen the best response.1
Quite alright, medals are not important.
 one year ago

Best_Mathematician Group TitleBest ResponseYou've already chosen the best response.0
exactly thx for explaining stuff
 one year ago
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