Quiz #1 Question #1


Moderators: Chem_Mod, Chem_Admin

Jacqui Dimalanta 3E
Posts: 4
Joined: Fri Jul 15, 2016 3:00 am

Quiz #1 Question #1

Postby Jacqui Dimalanta 3E » Fri Dec 02, 2016 8:20 pm

Hello. Just reviewing/correcting the old quizzes before the final lol. But, I need some help solving this problem:

Fluorescence is the property in which molecules absorb and emit light where the photon emitted is lower energy than the photon absorbed. The energy unaccounted for is dissipated across the molecule.
An adrenaline molecule was excited using 285 nm light. It dissipates 8.584 * 10^-20 J of energy per photon. What is the energy of light emitted in joules?

So, what I did was this.. (and correct me if I'm wrong)

Given: Change in E = 8.584 * 10^-20 J; Wavelength = 285 nm

I put 285 nm as 2.85 x 10^-7m into frequency = speed of light / wavelength and got v = 1.05 * 10^15. Then using E=hv, I got the energy at 285 nm to be 6.9573 * 10^-19 J.

And because we don't know the wavelength at the point the energy was emitted... I set up an equation like this:

8.584 * 10^-20 J = (frequency * 6.626 * 10^-34) - (6.9573 * 10^-19 J)

I found the frequency to be 1.180 * 10^13... then I multiplied it by Planck's constant to get 7.819 * 10^-21 J as the energy of light emitted

Not sure if this is right... though it does seem right since the final energy is lower.

404805762
Posts: 11
Joined: Wed Sep 21, 2016 2:58 pm

Re: Quiz #1 Question #1

Postby 404805762 » Sat Dec 03, 2016 5:05 pm

I'm pretty sure you did this right!

Alexandra_Cooper_4H
Posts: 10
Joined: Wed Sep 21, 2016 2:56 pm

Re: Quiz #1 Question #1

Postby Alexandra_Cooper_4H » Sun Dec 04, 2016 2:06 am

Hi,
I was partially confused on this problem, but what you did seems right. So I have a quick question, if you don't mind explaining your work.
Why did you subtract the energy you got at 285nm? Is it the change in energy= initial energy - final energy concept? I remember my TA writing that on my quiz but I couldn't figure out what he meant! What I do remember, is not being able to figure out how to calculate an emitted energy that was different from the change in energy we were given.
Thank you!

Jacqui Dimalanta 3E
Posts: 4
Joined: Fri Jul 15, 2016 3:00 am

Re: Quiz #1 Question #1

Postby Jacqui Dimalanta 3E » Sun Dec 04, 2016 11:19 am

Yes! But, it is actually change in energy = final energy - initial energy.

So, the change in energy of the photon was already given at 8.584 * 10^-20 J. Your final energy is the energy of light emitted, and the initial energy was found to be at a wavelength of 285 nm. And you use E=hv to find both the energies.

I'm thinking that the reason that the equation is set up like this is because the final energy is the energy the photon emits, and the initial is the energy absorbed. Subtracting them would give the change in energy by the photon.

Also, you can check your answer by looking at the problem... and it says that the photon emitted is at a lower energy than the photon absorbed. So that's how I felt my answer could be correct.

Hope this helped (:


Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest