Question

Photons help

Answer 1

use plancks equation:

E=hc/lambda

Answer 2

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

Answer 3

how?

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

so is the answer 1.24*10^-6

Answer 9

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

Answer 10

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

Answer 11

That's the wavelength :)

Answer 12

and the unit of wavelength is Hz right

Answer 13

Hertz is for frequency, Wavelength is meter,

Answer 14

3.89∗10^(−7) meters ???

Answer 15

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

Answer 16

oh ya thx...what abt 2nd one

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

haha thx tho

Answer 22

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

Answer 23

haha ok.

Answer 24

@Frostbite might know.

Answer 25

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

Answer 26

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

Answer 27

can't you use kirkchoffs laws here?

Answer 28

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.

Answer 29

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

Answer 30

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

Answer 31

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.

Answer 32

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
\[ \lambda = \frac{hc}{E_2-E_0} \]

Answer 33

is this first answer?

Answer 34

Yes.

Answer 35

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

Answer 36

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

Answer 37

oh gotcha

Answer 38

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?

Answer 39

so energy photons absorbed are all of them?

Answer 40

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

Answer 41

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

Answer 42

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.

Answer 43

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

Answer 44

No.

Answer 45

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

Answer 46

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

Answer 47

nice thanks...

Answer 48

wish to give u 3 medals but just have 1

Answer 49

Quite alright, medals are not important.

Answer 50

exactly thx for explaining stuff