Question

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $\lambda$ is $\text{V}_s$. If the intensity of the incident light is doubled while keeping wavelength identical, the stopping potential will be:

• A. \( 2\text{V}_s \)
• B. \( \frac{\text{V}_s}{2} \)
• C. \( \text{V}_s \)
• D. \( 4\text{V}_s \)

Hint:

Keep a crystal-clear mental separation between these two quantities in modern physics: Frequency/Wavelength controls the energy parameters (\(\text{K}_{\text{max}}\), stopping potential, individual photon speed), whereas Intensity controls the quantity parameters (number of photons, number of ejected electrons, total circuit current). Changing one leaves the other completely untouched!

Answer 1

The Correct Option is C

Concept: According to Einstein's Photoelectric Equation, the maximum kinetic energy (\(\text{K}_{\text{max}}\)) of an emitted photoelectron depends strictly on the energy of the incident photon (\(E\)) and the work function (\(\phi_0\)) of the target metal surface: \[ \text{K}_{\text{max}} = E - \phi_0 \] Expressing the incident photon energy in terms of its wavelength (\(\lambda\)): \[ \text{K}_{\text{max}} = \frac{hc}{\lambda} - \phi_0 \] The stopping potential (\(\text{V}_s\)) is the negative potential required to completely halt the fastest moving photoelectrons, meaning: \[ \text{K}_{\text{max}} = e\text{V}_s \quad \implies \quad e\text{V}_s = \frac{hc}{\lambda} - \phi_0 \] From this relation, it is clear that the stopping potential is governed purely by the frequency or wavelength of the incoming light and the nature of the metal surface. Intensity, by definition in quantum physics, is a measure of the number of photons striking a unit area per second. Changing the intensity modifies the total number of emitted electrons (the photocurrent) but has absolutely no effect on the individual energy configuration of each arriving photon.

Step 1: Analyzing the effect of changing the light intensity.
When the intensity of the incident light is doubled:
• The total number of incident photons striking the metal surface per second is doubled.
• Consequently, the total number of emitted photoelectrons per second (saturation photocurrent) also doubles.

Step 2: Evaluating the final stopping potential.
Because the problem specifies that the wavelength (\(\lambda\)) is kept completely identical, the energy per photon (\(E = \frac{hc}{\lambda}\)) remains unchanged. Since both the photon energy and the metal's work function are constant, the maximum kinetic energy of the individual photoelectrons remains completely unaffected: \[ \text{K}_{\text{max}}' = \text{K}_{\text{max}} \quad \implies \quad e\text{V}_s' = e\text{V}_s \] Therefore, the stopping potential remains entirely unchanged: \[ \text{V}_s' = \text{V}_s \]