1. Using the wave equation and the energy equation derive an equation that expresses a photon's energy in terms of its wavelength. What proportionality exists between energy and wavelength?

Question

Question

1. Using the wave equation and the energy equation derive an equation that expresses a photon's energy in terms of its wavelength. What proportionality exists between energy and wavelength?

Answer 1

### \( E = \frac{hc}{\lambda} \) and \( E \propto \frac{1}{\lambda} \)

Answer 2

## Step 1: Identify the relevant equations ### The wave equation for a photon is given by \( c = \lambda \nu \), where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( \nu \) is the frequency. The energy equation for a photon is \( E = h \nu \), where \( E \) is the energy and \( h \) is Planck's constant. ## Step 2: Express frequency in terms of wavelength ### From the wave equation, solve for frequency: \( \nu = \frac{c}{\lambda} \). ## Step 3: Substitute frequency into the energy equation ### Substitute \( \nu = \frac{c}{\lambda} \) into \( E = h \nu \) to get \( E = h \frac{c}{\lambda} \). ## Step 4: Simplify the energy equation ### The final equation expressing a photon's energy in terms of its wavelength is \( E = \frac{hc}{\lambda} \). ## Step 5: Determine the proportionality ### The energy \( E \) is inversely proportional to the wavelength \( \lambda \), as indicated by the equation \( E \propto \frac{1}{\lambda} \).

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1. Using the wave equation and the energy equation derive an equation that expresses a photon's energy in terms of its wavelength. What proportionality exists between energy and wavelength?

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