Question
Calculate the energy of a photon of a light with the frequency of 6.5e-14s
Answers
GPT-4o mini
To calculate the energy of a photon, you can use the formula:
\[
E = h \cdot f
\]
where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)),
- \( f \) is the frequency of the light (in hertz).
Given that the frequency \( f = 6.5 \times 10^{-14} \, \text{s}^{-1} \) (which should actually have units of hertz, or \( \text{s}^{-1} \)), we can plug in the values:
\[
E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \cdot (6.5 \times 10^{14} \, \text{s}^{-1})
\]
Now, calculate:
\[
E = 6.626 \times 10^{-34} \times 6.5 \times 10^{14}
\]
Calculating this gives:
\[
E \approx 4.303 \times 10^{-19} \, \text{J}
\]
Therefore, the energy of the photon is approximately \( 4.30 \times 10^{-19} \, \text{J} \).
\[
E = h \cdot f
\]
where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)),
- \( f \) is the frequency of the light (in hertz).
Given that the frequency \( f = 6.5 \times 10^{-14} \, \text{s}^{-1} \) (which should actually have units of hertz, or \( \text{s}^{-1} \)), we can plug in the values:
\[
E = (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \cdot (6.5 \times 10^{14} \, \text{s}^{-1})
\]
Now, calculate:
\[
E = 6.626 \times 10^{-34} \times 6.5 \times 10^{14}
\]
Calculating this gives:
\[
E \approx 4.303 \times 10^{-19} \, \text{J}
\]
Therefore, the energy of the photon is approximately \( 4.30 \times 10^{-19} \, \text{J} \).