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A $\ce{SO3}$ molecule is radiated with light with a wavelength of $\pu{193 nm}$, which results in photodissociation.

$\ce{SO3} + \text{photon}(\lambda=\pu{193nm})\ce{-> SO2 + O}$

It is measured that the $\ce{SO2}$ fragment is in the electronic ground state, but in the vibrational excited state where bending is in $v=3$.

  • Calculate the velocity of the $\ce{O}$ fragment after the dissociation.

I know that:

  • The $\ce{SO3}$ molecule is at rest before the radiation, and we don't have to look at the rotation.

  • The dissociation energy is $\pu{5.0 eV}$

  • The normal vibration frequencies for $\ce{SO2}$ is $\omega_1= \pu{1151 cm^{-1}}$ (sym stretch), $\omega_2= \pu{518 cm^{-1}}$ (bending) and $\omega_3= \pu{1362 cm^{-1}}$ (asym stretch)

My first thought was to approach it like the photoelectric effect where you can determine the velocity of the ejected electron. But I can't seem to figure out how to do it using the dissociation energy and vibrational modes. I hope someone can help me with this?

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    $\begingroup$ Your instinct is correct. Convert all units into a common energy unit. Then subtract the vibrational energy remaining in $\ce{SO2}$ from the incoming photon energy. The result is the kinetic energy of $\ce{O}$. Convert that into a velocity. $\endgroup$ – Buck Thorn Jul 14 at 20:44
  • $\begingroup$ What about the dissociation energy? Do I have to subtract that as well? $\endgroup$ – Simone Maarup Pedersen Jul 14 at 21:22
  • $\begingroup$ If I subtract it as well, I get the velocity to around 3510 m/s $\endgroup$ – Simone Maarup Pedersen Jul 14 at 21:33
  • $\begingroup$ Yes, of course, you also have to subtract the dissociation energy. $\endgroup$ – Buck Thorn Jul 15 at 6:58

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