Answer
(a) decreased
(b) $-10.23eV$
Work Step by Step
(a) We know that $E=\frac{hc}{\lambda}$. This equation shows that the energy of the photon is inversely proportional to its wavelength; therefore, the energy of the atom decreases.
(b) As $\Delta E=hc[\frac{1}{\lambda_{abs}}-\frac{1}{\lambda_{emi}}]$
We plug in the known values to obtain:
$\Delta E=(6.63\times 10^{-34}J.s)(3\times 10^8m/s)[\frac{1}{486.2\times 10^{-9}m}-\frac{1}{97.23\times 10^{-9}m}]$
$\Delta E=-16.3657\times 10^{-19}J$
$\Delta E=(-16.3657\times 10^{-19}J)(\frac{1eV}{1.6\times 10^{-19}J})=-10.23eV$
