Question

In a photoelectric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photoelectric current.
A. remains constant
B. is halved
C. is doubled
D. becomes four times

Answer 1

Hint:In order to answer this question, first we will explain the relation between the saturation photoelectric current and the intensity, frequency of the incident light, as saturation photoelectric current is frequency independent. And then we will discuss the whole concept of photoelectric current.

Complete step by step answer:
The saturation photoelectric current is proportional to the intensity of incident radiation, but it is frequency independent. As a result, when the intensity and frequency of the incident light are both doubled, the saturation photoelectric current is doubled.The photoelectric effect is a phenomenon in which light causes electrons to be expelled from a metal's surface. Photoelectrons are the electrons that are expelled. It's worth noting that the frequency of the light incident on the metal's surface affects the emission of photoelectrons and the kinetic energy of the expelled photoelectrons.

Photoemission is the term used to describe the process by which photoelectrons are emitted from the metal's surface due to the action of light. The photoelectric effect happens when electrons at the metal's surface absorb energy from incident light and use it to overcome the attraction forces that bind them to the nuclei of the metal.

The photoelectric effect cannot be explained using the wave model of light. However, the particle nature of light, which can be viewed as a stream of electromagnetic energy particles, can explain this behaviour. Photons are the ‘particles' of light that make up the visible spectrum. Planck's equation relates the energy stored by a photon to the frequency of light:
\[E = hv = \dfrac{{hc}}{\lambda }\]
where, $E$ denotes the energy of the proton, $h$ is Planck's constant, $v$ denotes the frequency of the light, $c$ is the speed of light in vacuum and $\lambda $ is the wavelength of the light.

Hence, the correct option is C.

Note:Further investigation demonstrated that "saturation current" is frequency independent. I couldn't find any information about instantaneous current (current other than saturation current). Speculation 1: If saturation current is not attained, higher frequency radiation will produce more photoelectric current.