Problem 22 - Chemistry: The Central Science

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?

Step-by-Step Solution

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 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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.

Step 5: 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: \(E_{IR} = (6.63 × 10^{-34} Js)(4 × 10^{14} Hz) \approx 2.65 × 10^{-19} J\)

Step 6: 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: \(E_{UV} = (6.63 × 10^{-34} Js)(1 × 10^{15} Hz) \approx 6.63 × 10^{-19} J\)

Step 7: 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.