What's the force experienced by a hydrogen atom when it absorbs photons? Suppose a hydrogen atom is a distance $d$ away from a star (approximate the star as a blackbody). The radius of the star is $R$ and the its temperature is $T$. The hydrogen atom absorbs photons from the star and goes from $1s$ to $2p$ state. It gains the photon's momentum, but in a short time $\Delta t$, the atom emits the photon isotropically, meaning the momentum change due to the emission of photons is $0$. I am now looking for the force due to the absorption of the photons.
We can use Stefan-Boltzmann's law to find the energy flux that goes through the hydrogen atom at that tiny sold angle. We can also find the force through the following equation,
\begin{equation}
    F = \frac{I}{\Delta t} = \frac{\Delta p}{\Delta t}
\end{equation}
where $I$ is the impulse and the $\Delta p$ is the change of momentum. It seems difficult to relate the energy flux to the change of momentum because only one specific frequency can excite the $1s$ state to $2p$. I am also unsure if we need to refer to the Einstein coefficient to find the transition rate.
 A: As I mentioned in the comments, the mechanism described in the OP is used in the laser cooling of atoms, and the detailed derivations of the force can be found by googling (although they may be complicated by taking into account the Doppler shift, which is not necessary here). I therefore make only a few points that are missing in the reasoning presented in the OP:

*

*Directionality the photons come from a certain direction, say with momenta $\mathbf{p}_i=\hbar\mathbf{k}_i$, but are reemitted in arbitrary directions.

*The rate of photon absorption and reemission the time of interaction of an atom and a photon is hard to define, i.e., one cannot really determine the force produced by one collision. On the other hand, one can calculate the net change of the momentum produced by many photons absorbed and reemitted, given that one knows how many such processes happen per second - here the Einstein coefficients could be helpful to relate this rate to the field intensity.

*Averaging As I have already mentioned, the photons are emitted in random directions. The changes of momenta transversal to $\mathbf{p}_i$ will average out after many collisions, leaving us with a net change (and hence the net force) only along the direction of the incident light.

A: The rate of photon absorption is $$ n = \int B_{12}I(\nu)\phi(\nu) d\nu\ ,$$
per atom, where $I(\nu)$ is the specific intensity, $B_{12}$ is the absorption Einstein coefficient (as defined by this equation and assuming stimulated emission is negligible) and $\phi(\nu)$ is a profile function, centred on the transition frequency, that must be integrated over, and which encapsulates any broadening processes.
Multiplying the integrand by an extra $h\nu/c$ will give the rate of change of momentum, or the force, per atom (assuming that spontaneous emission is isotropic). Multiplying by the number of atoms per unit volume gives the force on that volume.
This assumes that the gas is optically thin (i.e. the probability of a photon being absorbed at line centre is considerably less than 1). If it is optically thick in the line then a more complicated radiative transfer calculation will be needed and the force will be reduced.
