Why is sulfur dioxide more stable to UV radiation than ozone? Ozone readily decomposes when impacted by UV radiation.

Sulfur dioxide, having the same bond structure and valence electrons in each atom as ozone, is much more stable.  

Would someone please help me understand what makes sulfur dioxide a more stable compound?
I'm going to attempt to answer this from three points of view.
Qualitative considerations and MO theory:
First, why isn't ozone linear? You can imagine $\ce{O3}$ like a $\ce{CO2}$ molecule that lacks two electrons. This lack of electrons will result in less $s$-character, and less bonding &quot;attitude&quot;. That's why ozone is bent.
Still, the repulsion in its electrons ($\ce{O}$ is a small central atom, very small) will want it to be free. Oxygen's van der Waals radius is 1.5 $\mathrm{\overset{\circ}A}$. Thus, $\ce{O3}$ has a bond length of $\approx 1.2\ \mathrm{\overset{\circ}A}$. That's really cramped. Electrons aren't happy with this and will try to get free as soon as $E_a$ and other conditions are provided.
Now compare that with sulfur's 1.8 $\mathrm{\overset{\circ}A}$ radius; which, results in a $\approx 1.43\ \mathrm{\overset{\circ}A}$ bond length. Less cramped, the electrons aren't as much complaining as they do in $\ce{O3}$'s case. We can't get really further as these two molecules aren't isoelectronic.
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Approach using classic high school knowledge:
It's because oxygen has an arbitrarily positive formal charge. Take a look at $\ce{O3}$'s electron-dot diagram:
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The oxygen atom, which has an electronegativity of $3.6$ in Allen scale, has formed $sp^2$ orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very &quot;non-bonding&quot;.
The sulfur, on the other hand, has an electronegativity of $\approx 2.5$. It will also get a positive formal charge. But since it's less electronegative, it bears it.
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Thermodynamics and kinetics:
Ozone decays into diatomic oxygen under certain conditions:
$$\ce{2O3 -&gt; 3O2}$$
Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is $142~ \mathrm{kJ/mol}$, while it's 0 for diatomic oxygen. So
$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$
$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$
Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.
Above that, its energy of activation is a measly $\approx 32 \mathrm{~kJ/mol}$, which is a very small kinetic barrier.
Thus, you can also conclude that $\ce{O3}$ is exceptionally &quot;inclined to decompose&quot;. Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.