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This is kind of a coursework question but it bring up some really interesting things about the fine structure constant $\alpha$ so I wanted to post it to not only make sure I understood something but to get into some other ideas as well.

So, the question: we're asked to show that the ration of the wavelength of a photon emitted by an atom and its size is related in turn to $\alpha$. I came up with the following:

for a hydrogen-like atom (using the Bohr model) $$r = \frac{n^2 \hbar^2} {Zm_e k e^2}$$ or, in terms of the Bohr radius, $r= \frac{n^2r_0}{Z}$ where $k=\frac{1}{4\pi\epsilon_0}$

Now, the energy of an atom is $E=\frac{kZe^2}{2r}$ (I can get this from Virial theorem) and r in this case is the radius we have from earlier.

Plugging one into the other we have $$E=\frac{kZe^2}{2 \frac{n^2 \hbar^2} {Zm_e k e^2}}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2}$$ since $E=h \nu = hc/ \lambda$ we have $$\frac{hc}{\lambda}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2}$$

Now, the fine structure constant $\alpha = \frac{ke^2}{\hbar c}$ and that turns this into $$\frac{1}{\lambda}=\frac{k^2Z^2e^4 m_e}{2 n^2 \hbar^2 hc}=\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar} $$ and multiplying the original r by that: $$\frac{n^2 \hbar^2} {Zm_e k e^2}\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar}=\alpha \pi Z = \frac{r}{\lambda}$$

which tells me that the ratio of size to wavelength depends entirely on Z and $\alpha$.

I also plugged this into the change-of-energy equation,

$$\Delta E = -Z^2 \frac{ke^2}{2r_0} \left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)$$ and plugging in what we have for $r_0$:

$r_0=\frac{\hbar^2}{m_eke^2}\rightarrow-Z^2 \frac{ke^2}{2(\frac{\hbar^2}{m_eke^2})} \left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\rightarrow -Z^2\frac{k^2e^4m_e}{2\hbar^2}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)\rightarrow -Z^2\frac{\alpha^2c^2m_e}{2}\left(\frac{1}{n_i^2}-\frac{1}{n_f^2}\right)=\Delta E$

Now assuming I did this correctly, would the same apply to a molecule? That is, given a wavelength, I would think you'd just plug that back into the energy equation (using $\Delta E = \frac{hc}{\lambda} $ and get an estimate of the size. But I was checking if my logic was correct.

The other interesting thing to me is how one might use $\alpha$ in other intersting ways for problems like this.

Anyhow, if anyone can say that I did something dumb that's most appreciated. I just want to see if my logic is correct.

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You have: $$\frac{r}{\lambda} = \frac{n^2 \hbar^2} {Zm_e k e^2}\alpha^2 \frac{\pi Z^2 m_ec}{ n^2 \hbar}$$ $$=\alpha \pi Z c$$ I must have missed a step since I don't see how you cancelled out so many terms. Where did $\frac {\hbar}{ke^2}$ go to? –  LDC3 May 11 at 3:08
    
The $m_e$ goes away, $n^2$ goes, $Z^2$ becomes Z. That leaves $\frac{\hbar}{ke^2}\alpha^2\pi Z c$ which, since $\alpha=\frac{ke^2}{\hbar c}$ means it's $\alpha \pi Z$. I think that should work, but I may have messed it. :-) –  Jesse May 11 at 3:21
    
Oh, I found it, it should be: $$\frac{r}{\lambda} = \alpha \pi Z$$ since $\alpha = \frac{ke^2}{\hbar c}$ –  LDC3 May 11 at 3:22
    
yeah i fixed it. –  Jesse May 11 at 3:22
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up vote 3 down vote accepted

would the same apply to a molecule? That is, given a wavelength, I would think you'd just plug that back into the energy equation...and get an estimate of the size

No.

In the hydrogen atom, size is the distance of the electron from the proton.

In a molecule, size is primarily the relative locations of the nuclei. Especially for large molecules, absorption wavelengths are reflective of local structure, not the overall structure. For example a C=C bond will absorb at the about the same wavelength from one molecule to another.

To take an extreme example, consider a protein molecule.
The protein may contain 100s or 1000s of amino acid residues. But all protein molecule have strong absorbance in the 280nm range regardless of size.

An exception to the wavelength being indicative of only part of the molecule, would be if the entire molecule was a conjugated system have delocalized electrons.

In a system with an alternating series of double and single carbon-carbon bonds:

$$-C=C-C=C-C=C-C=C-$$ the wavelength of a transition will increase with the length of the series.

In the case of conjucated double bonds a particle in a box model is sometimes applied. See Conjugated Bonding in Cyanine Dyes: A "Particle In A Box" Model.

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Well, I wondered about that because the way the problem was phrased it said you could do a rough approximation by looking at a molecule with an emission of ~X nanometers (benzene in this case, from excited state to ground). –  Jesse May 11 at 3:24
    
benzene is a special case because it has delocalized electrons. I edited the answer to mention this exception. –  DavePhD May 11 at 11:24
    
thanks. that link is helpful. –  Jesse May 11 at 12:54
    
One other bit: so if I want to apply the PIB model to compound like that, and I know the wavelength of emission, I know the $\Delta E$ and then I know from solving a Schrödinger that $E=n^2\frac{h^2}{8mL}$ and solving for L gets me my bond ("box" length). Yes? –  Jesse May 11 at 14:40
    
the length would be more than one bond, it would be the length of the whole conjugated system. –  DavePhD May 11 at 21:18
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