To solve this problem, we need to use the formula for the energy of a photon, which is given by:
[tex]\[ E = h \cdot f \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant,
- [tex]\( f \)[/tex] is the frequency of the light.
Given values:
- Frequency ([tex]\( f \)[/tex]) of the gun's light: [tex]\( 4.91 \times 10^{14} \)[/tex] hertz,
- Planck's constant ([tex]\( h \)[/tex]): [tex]\( 6.61 \times 10^{-34} \)[/tex] joule-seconds.
Let's calculate the energy step-by-step.
1. Multiply the frequency [tex]\( f \)[/tex] by Planck's constant [tex]\( h \)[/tex]:
[tex]\[
E = 6.61 \times 10^{-34} \, \text{Joule-seconds} \times 4.91 \times 10^{14} \, \text{Hertz}
\][/tex]
2. Performing the multiplication:
[tex]\[
E \approx (6.61 \times 4.91) \times (10^{-34} \times 10^{14}) \, \text{Joules}
\][/tex]
[tex]\[
E \approx 32.43 \times 10^{-20} \, \text{Joules}
\][/tex]
3. Simplifying the expression:
[tex]\[
E \approx 3.2451 \times 10^{-19} \, \text{Joules}
\][/tex]
The value we found is approximately [tex]\( 3.2451 \times 10^{-19} \)[/tex] Joules.
Now, let's compare this value with the given options:
A. [tex]\( 3.24 \times 10^{-19} \)[/tex] Joules
B. [tex]\( 4.91 \times 10^{15} \)[/tex] Joules
C. [tex]\( 204 \times 10^{-21} \)[/tex] Joules
D. [tex]\( 4.92 \times 10^{-18} \)[/tex] Joules
The closest option to [tex]\( 3.2451 \times 10^{-19} \)[/tex] Joules is:
A. [tex]\( 3.24 \times 10^{-19} \)[/tex] Joules
Thus, the correct answer is:
A. [tex]\( 3.24 \times 10^{-19} \)[/tex] Joules.
