$(i)$ The energy $(E)$ of a photon is given by the expression:
$E = h \nu$
Where $h = 6.626 \times 10^{-34} \, J \cdot s$ (Planck's constant) and $\nu = 3 \times 10^{15} \, Hz$.
Substituting the values:
$E = (6.626 \times 10^{-34}) \times (3 \times 10^{15}) = 1.988 \times 10^{-18} \, J$.
$(ii)$ The energy $(E)$ of a photon with wavelength $(\lambda)$ is given by:
$E = \frac{hc}{\lambda}$
Where $h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^{8} \, m/s$,and $\lambda = 0.50 \, \mathring{A} = 0.50 \times 10^{-10} \, m$.
Substituting the values:
$E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{0.50 \times 10^{-10}} = 3.976 \times 10^{-15} \, J \approx 3.98 \times 10^{-15} \, J$.