Question E6F06:
Which of these materials is most commonly used to create photoconductive devices?  

Choices:  
A: Polyphenol acetate  
B: Argon  
C: Crystalline semiconductor  
D: All these choices are correct    

Correct Answer is  C: Crystalline semiconductor  

To understand photoconductive devices: their conductivity increases when illuminated because light generates electron-hole pairs in a semiconductor.

Why C (Crystalline semiconductor) is correct:
- Photoconductive cells (LDRs) are typically made from crystalline semiconductors such as cadmium sulfide (CdS), cadmium selenide (CdSe), lead sulfide (PbS), or silicon.
- In these materials, light creates mobile charge carriers, increasing conductivity and lowering resistance.

Why the others are incorrect:
- A: Polyphenol acetate — Not a standard photoconductive material. While some organic compounds can be photosensitive (e.g., in xerography), crystalline semiconductors are by far the most common for photoconductors used in electronics.
- B: Argon — An inert gas; not a photoconductive solid. It does not exhibit photoconductivity in the context of solid-state devices.
- D: All these choices are correct — False because Argon is not appropriate.

Useful formulas/rules of thumb:
- Conductivity increases with carrier concentration:
  $$\sigma = q(\mu_n n + \mu_p p)$$
  Under illumination, carrier densities increase:
  $$\Delta \sigma = q(\mu_n \Delta n + \mu_p \Delta p)$$
- Resistance is inversely related to conductivity:
  $$R \propto \frac{1}{\sigma}$$
- Photocurrent in a photoconductor approximately scales with optical power:
  $$I_{\text{ph}} \approx \eta \frac{q}{h\nu} P_{\text{opt}} \, G$$
  where $\eta$ is quantum efficiency, $G$ is photoconductive gain, and $h\nu$ is photon energy.

Rule of thumb: If the question mentions photoconductive cells or LDRs, think crystalline semiconductors like CdS.