Question

Most black-and-white photographic film (with the exception of some special-purpose films) is less sensitive to red light than blue light and has almost no sensitivity to infrared. How can these properties be understood on the basis of photons?

Short answer

The frequencies of blue, red and infrared photons decides teh occurrence of photoelectric effect.

Step-by-step solution

(a) Concept of photoelectric effect.

The photoelectric effect is the ejection of surface electrons in a metal when photon strikes. The mathematical expression governing the effect and giving the maximum kinetic energy of the ejected electrons is,

$$\begin{gathered} \mathit{K}\mathbf{.}{\mathit{E}}_{\mathbf{m}\mathbf{a}\mathbf{x}}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{\Phi } \\ \mathit{e}{\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{h}\mathit{f}\mathbf{-}\mathit{h}{\mathit{f}}_{\mathbf{0}} \end{gathered}$$

Where $\Phi$the work function, hf is the energy of the striking photon, f₀ is the threshold frequency and V₀ is the stopping potential.

(b) Explanation of the argument.

From the visible spectrum, the frequency of blue light is greater than the red light. It has to be noted also that the red light frequency is approximately closer or equal to the threshold frequency of a white-black photographic film. So, the red light photons are absorbed by the film’s electrons and thus the kinetic energy remains low when they are ejected.

The opposite happens for blue light photons where the threshold frequency is much lower than the frequency of the incident photons. Therefore, the ejected electrons have a greater kinetic energy to move.

For infrared photons, the frequency does not cross the threshold frequency and thus there will be no photoelectric effect and no ejected electrons.