To calculate the energy of a photon, we use the equation derived from Planck's equation:

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

where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant $(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})$,
• $c$ is the speed of light $(3.00 \times 10^8 \, \text{m/s})$,
• $\lambda$ is the wavelength of the photon in meters.

First, convert the given wavelength from nanometers (nm) to meters:

$\lambda = 715 \, \text{nm} = 715 \times 10^{-9} \, \text{m}$

Now, substitute the values into the energy equation:

$E = \frac{(6.626 \times 10^{-34}) \cdot (3.00 \times 10^8)}{715 \times 10^{-9}}$

Let me calculate this for you.It seems like the result was unexpectedly rounded to zero. Let me double-check and perform the calculation again carefully to ensure accurate results.The energy of the photon with a wavelength of 715 nm is approximately $2.78 \times 10^{-19}$ joules, rounded to three significant digits.

Do you have any further questions or need more details?

Here are some related questions for deeper understanding:

1. How does wavelength affect the energy of a photon?
2. What is the relationship between frequency and photon energy?
3. Can you calculate the energy of a photon with a shorter wavelength, like 500 nm?
4. How does Planck's constant influence photon energy calculations?
5. What are common applications of photon energy in everyday technology?

Tip: Energy and wavelength are inversely related, so shorter wavelengths result in higher energy photons.