# Determine the velocity of the O fragment after the dissociation

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?

• 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. – Buck Thorn Jul 14 '19 at 20:44
• What about the dissociation energy? Do I have to subtract that as well? – Simone Maarup Pedersen Jul 14 '19 at 21:22
• If I subtract it as well, I get the velocity to around 3510 m/s – Simone Maarup Pedersen Jul 14 '19 at 21:33
• Yes, of course, you also have to subtract the dissociation energy. – Buck Thorn Jul 15 '19 at 6:58