Question:
The following statement is correct in the case of photoelectric effect
The following statement is correct in the case of photoelectric effect
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In photoelectric effect:
\[
K_{\max}=h\nu-\phi
\]
Intensity controls photocurrent, while frequency controls maximum kinetic energy and stopping potential.
Updated On: Jun 22, 2026
- For a given frequency of incident radiation, the stopping potential varies linearly with its intensity.
- For a given frequency of incident radiation, the photocurrent is independent of its intensity.
- For a given frequency of incident radiation, the maximum kinetic energy of the photoelectrons is independent of its intensity.
- For a frequency lower than cutoff frequency, photoelectric emissions can occur if intensity of incident light is increased slightly.
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The Correct Option is C
Solution and Explanation
Step 1: Recall Einstein’s photoelectric equation.
According to Einstein’s photoelectric equation, \[ K_{\max}=h\nu-\phi \] where, \[ K_{\max}=\text{maximum kinetic energy of photoelectrons} \] \[ h=\text{Planck's constant} \] \[ \nu=\text{frequency of incident radiation} \] \[ \phi=\text{work function of the metal} \] Thus, the maximum kinetic energy depends only on frequency and not on intensity.
Step 2: Analyze option (1).
Stopping potential is related to maximum kinetic energy: \[ eV_0=K_{\max} \] Since \(K_{\max}\) depends on frequency and not on intensity, stopping potential also does not vary linearly with intensity.
Therefore, option (1) is incorrect.
Step 3: Analyze option (2).
Photocurrent depends on the number of emitted photoelectrons.
As intensity increases, more photons fall on the surface and more electrons are emitted. Hence photocurrent increases with intensity.
Therefore, option (2) is incorrect.
Step 4: Analyze option (4).
If the frequency of incident radiation is below threshold frequency, \[ \nu\lt \nu_0, \] then photoelectric emission cannot occur irrespective of intensity.
Increasing intensity only increases the number of photons, not the energy of each photon.
Therefore, option (4) is incorrect.
Step 5: Final conclusion.
Hence, the correct statement is: \[ \boxed{ \text{For a given frequency, the maximum kinetic energy is independent of intensity.} } \]
According to Einstein’s photoelectric equation, \[ K_{\max}=h\nu-\phi \] where, \[ K_{\max}=\text{maximum kinetic energy of photoelectrons} \] \[ h=\text{Planck's constant} \] \[ \nu=\text{frequency of incident radiation} \] \[ \phi=\text{work function of the metal} \] Thus, the maximum kinetic energy depends only on frequency and not on intensity.
Step 2: Analyze option (1).
Stopping potential is related to maximum kinetic energy: \[ eV_0=K_{\max} \] Since \(K_{\max}\) depends on frequency and not on intensity, stopping potential also does not vary linearly with intensity.
Therefore, option (1) is incorrect.
Step 3: Analyze option (2).
Photocurrent depends on the number of emitted photoelectrons.
As intensity increases, more photons fall on the surface and more electrons are emitted. Hence photocurrent increases with intensity.
Therefore, option (2) is incorrect.
Step 4: Analyze option (4).
If the frequency of incident radiation is below threshold frequency, \[ \nu\lt \nu_0, \] then photoelectric emission cannot occur irrespective of intensity.
Increasing intensity only increases the number of photons, not the energy of each photon.
Therefore, option (4) is incorrect.
Step 5: Final conclusion.
Hence, the correct statement is: \[ \boxed{ \text{For a given frequency, the maximum kinetic energy is independent of intensity.} } \]
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