[riboswitch]

In this problem, I'll assume that the electron particle is moving freely between two impenetrable barriers. 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!

[Corribus]

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.

[riboswitch]

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?

[Corribus]

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.

[riboswitch]

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. 

[Corribus]

Yes. (And thanks for using LaTex. I wish more posters took the trouble.)