Question

The following statement is correct in the case of photoelectric effect

• A. For a given frequency of incident radiation, the stopping potential varies linearly with its intensity.
• B. For a given frequency of incident radiation, the photocurrent is independent of its intensity.
• C. For a given frequency of incident radiation, the maximum kinetic energy of the photoelectrons is independent of its intensity.
• D. For a frequency lower than cutoff frequency, photoelectric emissions can occur if intensity of incident light is increased slightly.

Hint:

In photoelectric effect: \[ K_{\max}=h\nu-\phi \] Intensity controls photocurrent, while frequency controls maximum kinetic energy and stopping potential.

Answer 1

The Correct Option is C

Step 1: Recall Einstein's photoelectric equation.
Einstein explained the photoelectric effect using the concept of photons. Each photon of frequency $\nu$ carries energy $E = h\nu$. The maximum kinetic energy of emitted photoelectrons is: \[ K_{\max} = h\nu - \phi \] where $\phi$ is the work function (threshold energy) of the metal surface.
Step 2: Analyze option (1) -- stopping potential and intensity.
The stopping potential $V_0$ is related to $K_{\max}$ by $eV_0 = K_{\max} = h\nu - \phi$. Since $K_{\max}$ depends only on frequency $\nu$ and not on intensity, the stopping potential is also independent of intensity. So option (1) is INCORRECT.
Step 3: Analyze option (2) -- photocurrent and intensity.
Photocurrent is proportional to the number of photoelectrons emitted per unit time, which in turn is proportional to the number of incident photons per unit time (i.e., the intensity). So photocurrent IS dependent on intensity. Option (2) is INCORRECT.
Step 4: Analyze option (3) -- maximum KE and intensity.
From the Einstein equation $K_{\max} = h\nu - \phi$: the maximum kinetic energy depends only on the frequency $\nu$ and the work function $\phi$ of the metal. Intensity tells us how many photons arrive per second, not their individual energies. Therefore, $K_{\max}$ is completely independent of intensity. Option (3) is CORRECT.
Step 5: Analyze option (4) -- below cutoff frequency.
If the frequency of incident light is below the cutoff (threshold) frequency $\nu_0 = \phi/h$, then even a single photon does not have enough energy to eject an electron. Increasing intensity only increases the number of photons, not their individual energy. So photoelectric emission cannot occur below the cutoff frequency regardless of intensity. Option (4) is INCORRECT.
Step 6: State the correct answer.
For a given frequency of incident radiation, the maximum kinetic energy of photoelectrons is independent of the intensity of the incident light. \[ \boxed{\text{Option (3) is correct: } K_{\max} = h\nu - \phi \text{ is independent of intensity}} \]