Answer
The energy of the atom increases by $~~2.6~eV$
Work Step by Step
We can find the energy of a photon with a frequency of $6.2\times 10^{14}~Hz$:
$E = h~f$
$E = (4.136\times 10^{-15}~eV~s)(6.2\times 10^{14}~Hz)$
$E = 2.6~eV$
The energy of the atom increases by $~~2.6~eV$
