Answer
The net energy absorbed by the atom is $~~1.17~eV$
Work Step by Step
We can find the energy associated with $\lambda = 375~nm$:
$E = \frac{hc}{\lambda}$
$E = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{375\times 10^{-9}~m}$
$E = 3.309~eV$
We can find the energy associated with $\lambda = 580~nm$:
$E = \frac{hc}{\lambda}$
$E = \frac{(4.136\times 10^{-15}~eV~s)(3.0\times 10^8~m/s)}{580\times 10^{-9}~m}$
$E = 2.139~eV$
The atom absorbs $3.309~eV$ and emits $2.139~eV$
We can find the net energy gain:
$E = 3.309~eV-2.139~eV = 1.17~eV$
The net energy absorbed by the atom is $~~1.17~eV$
