To calculate the energy $E$ of a photon, we use the formula:

$E = \frac{h \cdot c}{\lambda}$

Where:

• $E$ is the energy of the photon (in Joules),
• $h$ is Planck's constant $= 6.626 \times 10^{-34} \, \text{J·s}$,
• $c$ is the speed of light in a vacuum $= 3.00 \times 10^8 \, \text{m/s}$,
• $\lambda$ is the wavelength of the photon $= 591.5 \, \text{nm} \, = 591.5 \times 10^{-9} \, \text{m}$.

Let's calculate it.

$E = \frac{(6.626 \times 10^{-34}) \times (3.00 \times 10^8)}{591.5 \times 10^{-9}}$

$E \approx 3.36 \times 10^{-19} \, \text{J}$

Thus, the energy of the photon is approximately $3.36 \times 10^{-19} \, \text{J}$.

Would you like more details or explanations on this?

Here are 5 related questions for further exploration:

1. How does the energy of a photon change with wavelength?
2. What is the energy of a photon with a wavelength of 450 nm?
3. How does the frequency of light relate to its energy?
4. What are the applications of photon energy calculations in real life?
5. How does the wavelength of visible light compare to other forms of electromagnetic radiation?

Tip: Photon energy decreases as wavelength increases, and vice versa.