## Best_Mathematician Group Title Photons help one year ago one year ago

1. aaronq Group Title

use plancks equation: E=hc/lambda

2. rohankv Group Title

Well use plank's equations.....however answer depends on geometry of molecule in case of part B & part C

3. Best_Mathematician Group Title

how?

4. .Sam. Group Title

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

5. Best_Mathematician Group Title

so is it 6.62*10^-34/3.2? hey what is c then?

6. .Sam. Group Title

"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, $\lambda=\frac{6.63 \times 10^{-3}(3 \times 10^8)}{3.2e}$

7. .Sam. Group Title

the charge, e, is $$1.6 \times 10^{-19}C$$

8. Best_Mathematician Group Title

9. .Sam. Group Title

How do you get that? should be $\lambda=\frac{6.63 \times 10^{-34}(3 \times 10^8)}{3.2(1.6 \times 10^{-19})}$

10. Best_Mathematician Group Title

ya and then i solved it using calculator...i got$3.89 * 10^{-7}$

11. .Sam. Group Title

That's the wavelength :)

12. Best_Mathematician Group Title

and the unit of wavelength is Hz right

13. .Sam. Group Title

Hertz is for frequency, Wavelength is meter,

14. Best_Mathematician Group Title

3.89∗10^(−7) meters ???

15. .Sam. Group Title

yes if you want to convert that into nanometers, then its $(3.89 \times 10^{-7}) \times 10^9 = 389nm$

16. Best_Mathematician Group Title

oh ya thx...what abt 2nd one

17. .Sam. Group Title

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

18. .Sam. Group Title

So the frequency plays a role here on the emission of photons.

19. .Sam. Group Title

I gotta go now, I'll leave it here, think about that :)

20. amistre64 Group Title

this is beyond my learning curve :) i can recall levers and rotations and such

21. Best_Mathematician Group Title

haha thx tho

22. abb0t Group Title

I would need to refresh my memory in modern physics. So no. Lol

23. Best_Mathematician Group Title

haha ok.

24. abb0t Group Title

@Frostbite might know.

25. Best_Mathematician Group Title

well looks like audience is here...but the players are not. haha

I will make some researches and see if i can help :3

27. abb0t Group Title

can't you use kirkchoffs laws here?

28. Jemurray3 Group Title

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.

29. Best_Mathematician Group Title

i really dont know how to do this...so if anyone can ex[lain it will be great

30. Best_Mathematician Group Title

i do think we r using planck's equation and stuff here

31. Jemurray3 Group Title

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.

32. Jemurray3 Group Title

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 $\lambda = \frac{hc}{E_2-E_0}$

33. Best_Mathematician Group Title

34. Jemurray3 Group Title

Yes.

35. Best_Mathematician Group Title

so the guy before who explained me the answer is wrong? we would need to include E2-E0 to get wavelength

36. Jemurray3 Group Title

Yes. His equations weren't wrong but you need the energy difference, not just the energy of the excited state.

37. Best_Mathematician Group Title

oh gotcha

38. Jemurray3 Group Title

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?

39. Best_Mathematician Group Title

so energy photons absorbed are all of them?

40. Jemurray3 Group Title

Photons could only be absorbed if that have E = (E2-E0), E = (E1-E0), or E= (E2-E1).

41. Best_Mathematician Group Title

ohk...so thats the second answer...what abt third one

42. Jemurray3 Group Title

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.

43. Best_Mathematician Group Title

oh so in last two answers we dont need any calculations and stuff?

44. Jemurray3 Group Title

No.

45. Best_Mathematician Group Title

what abt work function...is it the same thing as absorbing and emitting photons

46. Jemurray3 Group Title

No. The work function is not related to your question at all. It is a separate effect.

47. Best_Mathematician Group Title

nice thanks...

48. Best_Mathematician Group Title

wish to give u 3 medals but just have 1

49. Jemurray3 Group Title

Quite alright, medals are not important.

50. Best_Mathematician Group Title

exactly thx for explaining stuff