To calculate the energy of a photon given its frequency, 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]\(6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]),
- [tex]\( \nu \)[/tex] (the Greek letter "nu") is the frequency of the photon.
Given data:
- The frequency ([tex]\( \nu \)[/tex]) is [tex]\( 4.35 \times 10^{14} \, \text{Hz} \)[/tex].
Substitute the values into the equation:
[tex]\[ E = (6.63 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (4.35 \times 10^{14} \, \text{Hz}) \][/tex]
Performing the multiplication:
1. Multiply the numeric parts:
[tex]\[ 6.63 \times 4.35 = 28.8845 \][/tex]
2. Combine the powers of 10:
[tex]\[ 10^{-34} \times 10^{14} = 10^{-20} \][/tex]
So combining both steps:
[tex]\[ E = 28.8845 \times 10^{-20} \][/tex]
Convert this to scientific notation:
[tex]\[ 28.8845 \times 10^{-20} = 2.88845 \times 10^{-19} \][/tex]
Given the solution options and focusing on significant figures, we round the energy to three significant figures:
[tex]\[ E \approx 2.88 \times 10^{-19} \, \text{J} \][/tex]
Therefore, the energy of the photon is:
[tex]\[ \boxed{2.88 \times 10^{-19} \, \text{J}} \][/tex]
This matches the first option provided.
