Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: riboswitch on September 22, 2014, 02:56:00 PM
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In this problem, I'll assume that the electron particle is moving freely between two impenetrable barriers (http://en.wikipedia.org/wiki/Particle_in_a_box). It means I am solving a very simple "particle in a box" problem.
Problem:
The conjugated system of retinal consists of 11 carbon atoms and one oxygen atom. In the ground state of retinal, each level up to n=6 is occupied by two electrons. Assuming an average internuclear distance of 140 picometers, calculate:
- the separation in energy between the ground state and the first excited state in which one electron occupies the state with n=7;
- the frequency of the radiation required to produce a transition between these two states.
My answer:
I'm assuming that the particle is moving freely between two "barriers" distant 10 x 140 picometers (≈ 1 x 10-9 m). The quantized energies of an electron particle "in a box" are:
[tex]E_{n} = \frac{n^2h^2}{8mL^2} [/tex]
with
[tex]n = 1,2, .[/tex]
and L indicating the distance between the two barriers.
If I want to determine the amount of energy needed to "jump" from n=6 to n=7, then I have to calculate:
[tex]\Delta E_{6 \rightarrow 7} = E_{7} - E_{6}[/tex]
[tex]\Delta E_{6 \rightarrow 7} = \frac{13h^2}{8mL^2}[/tex]
[tex]\Delta E_{6 \rightarrow 7} = \frac{13 \times (6.626 \times 10^{-34} \ \frac{J}{s})^2 }{8 \ (9.11 \times 10^{-31}\ Kg) \ (1 \times 10^{-9} \ m)^2}[/tex]
Am I doing this well? Did I miss something? Did I do something wrong here? Is L okay?
Any input will be helpful. Thanks in advance!
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Approach seems OK to answer part 1. Didn't check any math. But you forgot to include the oxygen atom, which participates in the conjugation.
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So there are 10 carbon-carbon bonds and one carbon-oxygen bond.
We are going to pretend that each of these internuclear bonds have a length of 140 picometers.
So 11 x 140 picometers ≈ 1 x 10-9 meters, which was the value I used in my calculations. ;)
So:
[tex]\Delta E_{6 \rightarrow 7} = \frac{13 \times (6.626 \times 10^{-34} \ \frac{J}{s})^2 }{8 \ (9.11 \times 10^{-31}\ Kg) \ (1 \times 10^{-9} \ m)^2}[/tex]
Then I'll just use my calculator to get the value of ΔE.
I hope I'm doing this correctly.
Approach seems OK to answer part 1.
Oh, I totally forgot question number 2!
Let's see... I know that:
[tex]\Delta E \ = h \nu [/tex]
To calculate the frequency, I simply have to use this formula:
[tex]\nu \ = \frac{ \Delta E_{6 \rightarrow 7}}{h}[/tex]
Is this approach correct?
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Well, I couldn't necessarily call 1.54 x 10-9 meters "approximately" 1 x 10-9 meters. And without including the oxygen, the box length is 1.4 x 10-9 meters. You will get very different answers for all these values. Converting into light wavelengths, including oxygen in the conjugation path (12 total adjacent nuclei) gives me a transition wavelength of ~ 600 nm (orange light). Neglecting the oxygen (11 total adjacent nuclei) gives a transition wavelength at ~497 nm (blue-green). Using your approximation of 1 nm for the box length, the projected transition wavelength is 254 nm, somewhere in the UV. The experimental absorption maximum of rhodopsin, which utilizes retinal as a cofacter, is about 570 nm. So you see, this all does actually make a difference. Especially here, where the box length is squared, small approximations can give rise to large deviations... and approximating 1.5 as 1.0 is anything but small.
Part 2 looks OK.
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Well, I couldn't necessarily call 1.54 x 10-9 meters "approximately" 1 x 10-9 meters. And without including the oxygen, the box length is 1.4 x 10-9 meters. You will get very different answers for all these values. Converting into light wavelengths, including oxygen in the conjugation path (12 total adjacent nuclei) gives me a transition wavelength of ~ 600 nm (orange light). Neglecting the oxygen (11 total adjacent nuclei) gives a transition wavelength at ~497 nm (blue-green). Using your approximation of 1 nm for the box length, the projected transition wavelength is 254 nm, somewhere in the UV. The experimental absorption maximum of rhodopsin, which utilizes retinal as a cofacter, is about 570 nm. So you see, this all does actually make a difference. Especially here, where the box length is squared, small approximations can give rise to large deviations... and approximating 1.5 as 1.0 is anything but small.
Part 2 looks OK.
Now I know why I'm doing these exercises the wrong way.
I should abandon my habit of approximating the length of the "box".
So the new equation should look like this:
[tex]\Delta E_{6 \rightarrow 7} = \frac{13 \times (6.626 \times 10^{-34} \ \frac{J}{s})^2 }{8 \ \times (9.11 \times 10^{-31}\ Kg) \ \times (1.54 \times 10^{-9} \ m)^2}[/tex]
Doing the math will give me:
[tex]\Delta E_{6 \rightarrow 7} = 3.302 \times \ 10^{-19} \ J[/tex]
And the corresponding frequency is:
[tex] \nu \ \approx \ 4.9836 \times 10^{14} \ Hz[/tex]
You have already stated pretty much everything else. Thank you. ;D
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Yes. (And thanks for using LaTex. I wish more posters took the trouble.)