Answer :
To determine the energy of a photon given its frequency and Planck's constant, we use the formula for energy of a photon:
[tex]\[ E = h \cdot f \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.63 \times 10^{-34} \)[/tex] Joule-seconds),
- [tex]\( f \)[/tex] is the frequency of the photon ([tex]\(7.3 \times 10^{-17}\)[/tex] Hertz).
First, we substitute the given values into the formula:
[tex]\[ E = (6.63 \times 10^{-34}) \cdot (7.3 \times 10^{-17}) \][/tex]
Perform the multiplication:
[tex]\[ E = 6.63 \times 7.3 \times 10^{-34} \times 10^{-17} \][/tex]
[tex]\[ E = 48.399 \times 10^{-51} \][/tex]
Next, we simplify the result to match the form [tex]\( a \times 10^b \)[/tex], aiming for [tex]\(10^{-50}\)[/tex]:
[tex]\[ E = 4.8399 \times 10^{-50} \][/tex]
To express this value to the nearest tenth place, we look at the first decimal place and round accordingly:
[tex]\[ 4.8399 \approx 4.8 \][/tex]
Therefore, the energy of the photon is:
[tex]\[ 4.8 \times 10^{-50} \][/tex] Joules.
[tex]\[ E = h \cdot f \][/tex]
where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.63 \times 10^{-34} \)[/tex] Joule-seconds),
- [tex]\( f \)[/tex] is the frequency of the photon ([tex]\(7.3 \times 10^{-17}\)[/tex] Hertz).
First, we substitute the given values into the formula:
[tex]\[ E = (6.63 \times 10^{-34}) \cdot (7.3 \times 10^{-17}) \][/tex]
Perform the multiplication:
[tex]\[ E = 6.63 \times 7.3 \times 10^{-34} \times 10^{-17} \][/tex]
[tex]\[ E = 48.399 \times 10^{-51} \][/tex]
Next, we simplify the result to match the form [tex]\( a \times 10^b \)[/tex], aiming for [tex]\(10^{-50}\)[/tex]:
[tex]\[ E = 4.8399 \times 10^{-50} \][/tex]
To express this value to the nearest tenth place, we look at the first decimal place and round accordingly:
[tex]\[ 4.8399 \approx 4.8 \][/tex]
Therefore, the energy of the photon is:
[tex]\[ 4.8 \times 10^{-50} \][/tex] Joules.
