Question

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?

Short answer

By calculating the energy of a single photon for both infrared radiation \(E_{IR}\) and ultraviolet radiation \(E_{UV}\) using Planck's equation, we find that \(E_{IR} \approx 2.65 × 10^{-19} J\) and \(E_{UV} \approx 6.63 × 10^{-19} J\). Comparing these values, we conclude that ultraviolet radiation yields more electrical energy on a per-photon basis than infrared radiation, assuming equal efficiency of conversion in photovoltaic cells.

Step-by-step solution

List given information

We are given that the conversion efficiency is the same for both infrared and ultraviolet radiation. We need to compare the energy of a single photon of each radiation type.

Planck's Equation

As mentioned in the analysis, we can determine the energy of a photon using Planck's equation: \(E = h\nu\) Here, E is the energy of the photon, h is the Planck's constant (approximately \(6.63 × 10^{-34} Js\)), and ν is the frequency.

Relate frequency to the wavelength

The frequency (ν) and wavelength (λ) of a photon are related by their product to the speed of light, denoted by 'c'. The following formula can be used to find the frequency: \(\nu = \frac{c}{\lambda}\) Here, c is the speed of light, approximately \(3 × 10^8 m/s\), and λ is the wavelength in meters.

Compare energy for infrared and ultraviolet radiation

Now we need to determine the energy of a single photon of each type of radiation and compare them. Infrared radiation has a wavelength range of approximately \(700 nm\) to \(1 mm\), while ultraviolet radiation has a wavelength range of approximately \(10 nm\) to \(400 nm\). We can substitute the wavelength into the formula above and find the energy of a photon for both infrared and ultraviolet radiations. Since we are comparing the energy for a single photon, it will suffice to use the typical wavelength values for each radiation type.

Energy for a single infrared photon

For infrared radiation, let’s choose a wavelength of \(750 nm\). First, convert the wavelength to meters: \(750 nm = 750 × 10^{-9} m\) Now, using the frequency formula: \(\nu_{IR} = \frac{3 × 10^8 m/s}{750 × 10^{-9} m} \approx 4 × 10^{14} Hz\) Using the Planck's equation: \(E_{IR} = (6.63 × 10^{-34} Js)(4 × 10^{14} Hz) \approx 2.65 × 10^{-19} J\)

Energy for a single ultraviolet photon

For ultraviolet radiation, let’s choose a wavelength of \(300 nm\). First, convert the wavelength to meters: \(300 nm = 300 × 10^{-9} m\) Now, using the frequency formula: \(\nu_{UV} = \frac{3 × 10^8 m/s}{300 × 10^{-9} m} \approx 1 × 10^{15} Hz\) Using the Planck's equation: \(E_{UV} = (6.63 × 10^{-34} Js)(1 × 10^{15} Hz) \approx 6.63 × 10^{-19} J\)

Compare the photon energies

Comparing the energy of a single photon of infrared radiation \(E_{IR} \approx 2.65 × 10^{-19} J\) to the energy of a single photon of ultraviolet radiation \(E_{UV} \approx 6.63 × 10^{-19} J\), we can determine that ultraviolet radiation yields more electrical energy on a per-photon basis than infrared radiation, assuming equal efficiency of conversion in photovoltaic cells.