Does water undergo thermolysis if a nuclear device detonates within it? I caught a glimpse of the Wahoo nuclear test, an underwater test as part of Operation Hardtack I, and I wondered; that's a LOT of thermal energy going directly into the surrounding water, with this vast amount of thermal energy it's reasonable to say that some of the water underwent thermolysis, splitting it into hydrogen and oxygen gasses.
Is this assumption correct? If so how can we calculate what portion of the water underwent thermolysis? Finally, if this assumption is true, would the lysed hydrogen and oxygen recombine through combustion and the heat of the explosion?
 A: The temperature in a nuclear explosion is around
$T\approx 10^8\text{ K}$.
This results in an average kinetic energy per particle of
$$E=kT
 = 1.4\cdot 10^{-23}\text{ J/K} \cdot 10^8\text{ K}
 = 1.4\cdot 10^{-15}\text{ J}
 \approx 9000\text{ eV}$$
where $k$ is Boltzmann's constant and $\text{eV}$
(electronvolt) is a convenient unit of energy used in chemistry
and atomic physics.
This large energy ($E\approx 9000\text{ eV}$) is more than enough
to thermolyze the $H_2O$ molecules to $H$ and $O$ atoms,
because this chemical reaction needs only a few $\text{eV}$.
(Remember, all chemical bond energies are typically
in the range between $100\ ...\ 1000\text{ kJ/mol}$,
or equivalently $1\ ...\ 10\text{ eV}$ per molecule.)
Moreover, this large energy ($E\approx 9000\text{ eV}$) is also
more than enough to fully ionize the $H$ and $O$ atoms,
because the binding energies of the electrons are much lower
($13.6\text{ eV}$ for the single electron of the $H$ atom,
and around $500\text{ eV}$ for the innermost electrons
of the $O$ atom).
Hence a nuclear explosion in water will produce
a fully ionized plasma consisting of $H$ nuclei
(i.e. protons), $O$ nuclei, and electrons.
