To determine the energy of a photon, we use the formula from quantum mechanics which relates the energy [tex]\( E \)[/tex] of a photon to its frequency [tex]\( f \)[/tex]:
[tex]\[ E = hf \][/tex]
Here:
- [tex]\( h \)[/tex] is Planck's constant, valued at [tex]\( 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex].
- [tex]\( f \)[/tex] is the frequency of the photon, given as [tex]\( 3.6 \times 10^{15} \, \text{Hz} \)[/tex].
Let's perform the calculation step-by-step:
1. Substitute the values into the equation:
[tex]\[
E = (6.63 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (3.6 \times 10^{15} \, \text{Hz})
\][/tex]
2. Write out the multiplication:
[tex]\[
E = 6.63 \times 3.6 \times 10^{-34} \times 10^{15} \, \text{J}
\][/tex]
3. Multiply the constants:
[tex]\[
6.63 \times 3.6 = 23.868
\][/tex]
4. Combine the exponents of 10:
[tex]\[
10^{-34} \times 10^{15} = 10^{-19}
\][/tex]
5. Combine the results:
[tex]\[
E = 23.868 \times 10^{-19} \, \text{J}
\][/tex]
6. Convert the answer into scientific notation (keeping two significant figures, as the given choices are in this format):
[tex]\[
23.868 \times 10^{-19} = 2.3868 \times 10^{-18} \, \text{J}
\][/tex]
After reviewing the significant figures and matching it to the closest option given in the possible answers, we find the correct option:
[tex]\[ 2.4 \times 10^{-18} \, \text{J} \][/tex]
Thus, the energy of a photon with a frequency of [tex]\( 3.6 \times 10^{15} \, \text{Hz} \)[/tex] is:
[tex]\[ \boxed{2.4 \times 10^{-18} \, \text{J}} \][/tex]
