Answer
The smallest energy photon that can be absorbed has an energy of $10.2~eV$
Work Step by Step
The smallest energy photon would have the energy to make the electron jump from the ground state to the n=2 state.
We can find the energy when the atom is in the n=2 state:
$E = \frac{E_0}{n^2}$
$E = \frac{-13.6~eV}{2^2}$
$E = -3.4~eV$
We can find the energy difference between the two states:
$\Delta E = E_f-E_i$
$\Delta E = (-3.4~eV)-(-13.6~eV)$
$\Delta E = 10.2~eV$
The smallest energy photon that can be absorbed has an energy of $10.2~eV$.
