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Answer :
Certainly! Let's solve the first question, which is to calculate the energy of a photon with a frequency of [tex]\(5.0 \times 10^{14} \, \text{Hz}\)[/tex].
To find the energy of a photon, we use the formula that relates energy ([tex]\(E\)[/tex]) to frequency ([tex]\(f\)[/tex]):
[tex]\[ E = h \cdot f \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon.
- [tex]\( h \)[/tex] is Planck's constant, which is approximately [tex]\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex].
- [tex]\( f \)[/tex] is the frequency of the photon.
Now, substituting the given values into the equation:
[tex]\[ E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (5.0 \times 10^{14} \, \text{Hz}) \][/tex]
When you multiply these values together, you calculate the energy of the photon to be approximately:
[tex]\[ E \approx 3.313 \times 10^{-19} \, \text{J} \][/tex]
This means the energy of a photon with a frequency of [tex]\(5.0 \times 10^{14} \, \text{Hz}\)[/tex] is approximately [tex]\(3.313 \times 10^{-19} \, \text{J}\)[/tex].
To find the energy of a photon, we use the formula that relates energy ([tex]\(E\)[/tex]) to frequency ([tex]\(f\)[/tex]):
[tex]\[ E = h \cdot f \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon.
- [tex]\( h \)[/tex] is Planck's constant, which is approximately [tex]\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)[/tex].
- [tex]\( f \)[/tex] is the frequency of the photon.
Now, substituting the given values into the equation:
[tex]\[ E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \times (5.0 \times 10^{14} \, \text{Hz}) \][/tex]
When you multiply these values together, you calculate the energy of the photon to be approximately:
[tex]\[ E \approx 3.313 \times 10^{-19} \, \text{J} \][/tex]
This means the energy of a photon with a frequency of [tex]\(5.0 \times 10^{14} \, \text{Hz}\)[/tex] is approximately [tex]\(3.313 \times 10^{-19} \, \text{J}\)[/tex].
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