An atom absorbs a photon of wavelength 500nm and emits another...

The energy of a photon is given by $E = \dfrac{hc}{\lambda}$.

It is given that ${\lambda}_{1} = 500nm$, ${\lambda}_{2} = 600nm$.

The net energy absorbed is

$\DeltaE = \dfrac{hc}{{\lambda}_{1}} - \dfrac{hc}{{\lambda}_{2}} = \dfrac{hc}{{10}^{- 9}} ( \dfrac{1}{500} - \dfrac{1}{600} )$

$= \dfrac{6 . 6 \times {10}^{- 34} \times 3 \times {10}^{8} \times 100}{500 \times 600 \times {10}^{- 9}}$

$= \dfrac{6 . 6 \times 3}{30} \times {10}^{- 19}J$

$1eV = 1 . 6 \times {10}^{- 19}J$

$= \dfrac{6 . 6 \times 3}{30 \times 1 . 6}eV$

$= 4125 \times {10}^{- 4}eV$