To determine the energy of a photon emitting light with a frequency of [tex]\(4.47 \times 10^{14}\)[/tex] Hz, we can use Planck's equation:
[tex]\[ E = h \nu \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.62607015 \times 10^{-34}\)[/tex] Js),
- [tex]\( \nu \)[/tex] (or [tex]\( f \)[/tex]) is the frequency of the light.
Given:
- Frequency, [tex]\( \nu = 4.47 \times 10^{14} \)[/tex] Hz,
- Planck's constant, [tex]\( h = 6.62607015 \times 10^{-34} \)[/tex] Js,
The energy of the photon can be calculated as follows:
[tex]\[ E = (6.62607015 \times 10^{-34} \, \text{Js}) \times (4.47 \times 10^{14} \, \text{Hz}) \][/tex]
[tex]\[ E = 2.96185335705 \times 10^{-19} \, \text{J} \][/tex]
Therefore, the energy of a photon that emits light of frequency [tex]\(4.47 \times 10^{14}\)[/tex] Hz is approximately [tex]\(2.96 \times 10^{-19} \, \text{J}\)[/tex], which corresponds to option C.
So, the correct answer is:
C. [tex]\(2.96 \times 10^{-19} \, \text{J}\)[/tex]
