How is the size calculated of the area near a hydrogen atom that a photon must hit to ionize the atom? I do not know of any specific other information needed to answer the question.
 A: This lesson from Richard Fitzpatrick’s Quantum Mechanics course calculates the cross section for a hydrogen atom in its ground state to be ionized by an incoming photon. The result (eq. 8.259) is
$$\sigma\approx\frac{256\pi}{3}\alpha\left(\frac{I}{\hbar\omega}\right)^{7/2}\left(1-\frac{I}{\hbar\omega}\right)^{3/2}a_0^2 $$
where $a_0$ is the Bohr radius, $\alpha$ is the fine-structure constant, $\hbar$ is the reduced Planck constant, $\omega$ is the angular frequency of the photon, and $I=\hbar^2/(2m_ea_0^2)$ is the ionization energy of the ground state of hydrogen. (In that last formula, $m_e$ is the mass of the electron.)
This formula is valid when the photon’s energy is much greater than the ionization energy ($\hbar\omega\gg I$) and the ionized electron is non-relativistic. (The linked page has references to more complicated calculations when these constraints don’t apply.)
In this regime, the cross section is much smaller than the size of the atom. For example, when the photon has 100 times the ionization energy ($\hbar\omega/I=100$), $\sigma=1.9\times 10^{-7}a_0^2$. This is about $5.4\times 10^{-28}\text{ m}^2.$
The calculation of the cross-section involves applying time-dependent perturbation theory to the hydrogen ground state found by solving the Schrodinger equation for hydrogen. The initial ground state gets perturbed by the time-varying electromagnetic field of the photon, and there is some probability that the final state of the atom is an ionized state.
