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anonymous
 3 years ago
Photons help
anonymous
 3 years ago
Photons help

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aaronq
 3 years ago
Best ResponseYou've already chosen the best response.0use plancks equation: E=hc/lambda

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well use plank's equations.....however answer depends on geometry of molecule in case of part B & part C

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0They 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is it 6.62*10^34/3.2? hey what is c then?

.Sam.
 3 years ago
Best 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, \[\lambda=\frac{6.63 \times 10^{3}(3 \times 10^8)}{3.2e}\]

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0the charge, e, is \(1.6 \times 10^{19}C\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so is the answer 1.24*10^6

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ya and then i solved it using calculator...i got\[3.89 * 10^{7}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and the unit of wavelength is Hz right

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0Hertz is for frequency, Wavelength is meter,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.03.89∗10^(−7) meters ???

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ya thx...what abt 2nd one

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0The 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

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0So the frequency plays a role here on the emission of photons.

.Sam.
 3 years ago
Best ResponseYou've already chosen the best response.0I gotta go now, I'll leave it here, think about that :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0this is beyond my learning curve :) i can recall levers and rotations and such

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0I would need to refresh my memory in modern physics. So no. Lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well looks like audience is here...but the players are not. haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I will make some researches and see if i can help :3

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.0can't you use kirkchoffs laws here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0A 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i really dont know how to do this...so if anyone can ex[lain it will be great

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i do think we r using planck's equation and stuff here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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 \[ \lambda = \frac{hc}{E_2E_0} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this first answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the guy before who explained me the answer is wrong? we would need to include E2E0 to get wavelength

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yes. His equations weren't wrong but you need the energy difference, not just the energy of the excited state.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so energy photons absorbed are all of them?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Photons could only be absorbed if that have E = (E2E0), E = (E1E0), or E= (E2E1).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohk...so thats the second answer...what abt third one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh so in last two answers we dont need any calculations and stuff?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what abt work function...is it the same thing as absorbing and emitting photons

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No. The work function is not related to your question at all. It is a separate effect.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wish to give u 3 medals but just have 1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Quite alright, medals are not important.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exactly thx for explaining stuff
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