(a) An atom initially in an energy level with E = -6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with E = -2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?

According to de Broglie, Light has dual nature as in some situations it behaves like waves and in others like particles. Electrons, which sometimes exhibits particle nature may also behave like wave in some situations.

When a particle acts like a wave, it carries a wavelength which is known as de Broglie wavelength.

The energy of photon absorbed is

$$\begin{gathered} E=\frac{hc}{\lambda } \\ =\frac{\left(4.1357*{10}^{-15}eVs\right)\left(3*{10}^{8}m/s\right)}{860*{10}^{-19}m} \\ =1.4417eV \end{gathered}$$

The internal energy of the atom will be increased by this amount.

$E_{atom}=-6.52eV+1.4417eV=-5.078$

Therefore, the internal energy of the atom after absorption is -5.078 eV.

The energy of photon emitted is

$$\begin{gathered} E=\frac{hc}{\lambda } \\ =\frac{\left(4.1357*{10}^{-15}eVs\right)\left(3*{10}^{8}m/s\right)}{420*{10}^{-19}m} \\ =2.952eV \end{gathered}$$

The internal energy of the atom will be decreased by this amount.

$E_{atom}=-2.68eV+2.952eV=-5.632eV$

Therefore, the internal energy of the atom after emission is -5.632 eV.