Step 1 : Understanding the Question:
This question asks us to identify how the stopping potential of photoelectrons changes when the intensity of the incident light is doubled, while keeping its wavelength constant. The photoelectric effect describes the emission of electrons when light shines on a material. We need to understand whether the energy of the emitted photoelectrons is governed by the intensity of the light source or by its frequency and wavelength.
Step 2 : Key Formulas and Approach:
The relationship between the incident light energy and the kinetic energy of the emitted electrons is described by Einstein's photoelectric equation:
\[ K_{\text{max}} = e V_s = h \nu - \phi_0 \]
In terms of the wavelength of the incident light, the equation can be rewritten as:
\[ e V_s = \frac{h c}{\lambda} - \phi_0 \]
Here, $K_{\text{max}}$ represents the maximum kinetic energy of the emitted photoelectrons, $e$ is the elementary charge, $V_s$ is the stopping potential, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength of the incident light, and $\phi_0$ is the work function of the metal.
We will analyze the influence of light intensity on the parameters of this equation to find the new stopping potential.
Step 3 : Detailed Explanation:
According to wave-particle duality, light consists of packets of energy called photons. The energy of each individual photon depends solely on its frequency or wavelength, as given by $E = \frac{h c}{\lambda}$.
The intensity of light is a measure of the number of photons striking a unit area of the surface per unit time. Increasing the intensity at a constant wavelength increases the flux of photons, but does not alter the energy carried by any individual photon.
When the intensity of the light is doubled, the number of photons striking the metal surface per second is doubled. As a result, the number of photoelectrons ejected per second (and hence the saturation photocurrent) is also doubled, provided the frequency of the light is above the threshold frequency.
However, because the wavelength $\lambda$ is kept completely identical, the energy of each incident photon remains exactly the same. Since the work function $\phi_0$ of the metal is also constant, the maximum kinetic energy $K_{\text{max}}$ of the individual photoelectrons remains unchanged.
Because the maximum kinetic energy of the photoelectrons does not change, the stopping potential required to halt these electrons, defined by $V_s = \frac{K_{\text{max}}}{e}$, also remains completely unchanged.
Therefore, doubling the intensity of the incident light has no effect on the stopping potential, and it remains equal to the original value $V_s$.
Step 4 : Final Answer:
The stopping potential of the photoelectrons remains unchanged at $V_s$, which corresponds to Option (C).