Which statement describes a lead sulfide cell's sensitivity to light?

• A. Infrared rays
• B. Visible light
• C. Ultraviolet rays
• D. X-rays

Answer: (A)

The main idea is that a photodetector responds to light based on the semiconductor’s bandgap. A photon can free an electron only if its energy is at least the bandgap energy. Lead sulfide has a very small bandgap (about 0.37 eV at room temperature), which corresponds to wavelengths around 3 micrometers—squarely in the infrared region. That means PbS detectors are most responsive to infrared photons, because those wavelengths have just enough energy to bridge the gap and generate a measurable photocurrent. Higher-energy photons from ultraviolet or X-ray regions don’t define the detector’s primary sensitivity, and while some visible photons could interact, the device is optimized for infrared response, so infrared rays describe its sensitivity best.