It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

Planck's equation relates the energy of a photon (E) to its frequency (v) through the proportionality constant Planck's constant (h). The equation is given by: \[ E = h \times v\] where h = 6.63 × 10^{-34} Js (Planck's constant).

The frequency (v) of a photon can also be expressed in terms of its speed (c) and wavelength (λ) using the formula: \[ v = \cfrac{c}{\lambda}\] Knowing the numerical values of speed of light (c), we can calculate and compare the frequencies for infrared and ultraviolet radiation based on their wavelengths.

Now that we have the formulas connecting the photon's energy, frequency, and the speed of light, the comparison between the energy of infrared and ultraviolet photons becomes straightforward. 1. Infrared (IR) radiation has a longer wavelength (typically 700 nm to 1 mm) while Ultraviolet (UV) radiation has a shorter wavelength (typically 10 nm to 400 nm). 2. Shorter wavelengths result in higher frequencies and vice versa - this means that the UV radiation has higher frequency compared to the IR radiation. 3. Using Planck's equation, as the frequency of the photons increases, so does the energy of the photons. 4. Since UV photons have a higher frequency, they have more energy per photon compared to IR photons. From this analysis, we can conclude that ultraviolet radiation would yield more electrical energy on a per-photon basis than infrared radiation, assuming equal efficiency of conversion.