Question:
The kinetic energy of the released electron in photoelectric effect depends on
The kinetic energy of the released electron in photoelectric effect depends on
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Remember:
\[
K_{\max}=h\nu-\phi.
\]
\[
\text{Frequency } \rightarrow \text{controls kinetic energy}
\]
\[
\text{Intensity } \rightarrow \text{controls photoelectric current}
\]
This is one of the most important results of Einstein's photoelectric theory.
Updated On: Jul 29, 2026
- Intensity of incident photons
- Frequency of incident photons
- Area of photocell
- Time
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The Correct Option is B
Solution and Explanation
Concept:
According to Einstein's photoelectric equation,
\[
K_{\max}=h\nu-\phi,
\]
where
\[
K_{\max}=\text{maximum kinetic energy of photoelectrons},
\]
\[
h=\text{Planck's constant},
\]
\[
\nu=\text{frequency of incident radiation},
\]
\[
\phi=\text{work function of the metal}.
\]
Step 1: Observe the factors affecting kinetic energy. From \[ K_{\max}=h\nu-\phi, \] the kinetic energy depends directly on the frequency \(\nu\) of the incident photons.
Step 2: Examine the effect of intensity. Intensity controls the number of emitted photoelectrons (photoelectric current), but does not affect their maximum kinetic energy. \[ \boxed{\text{Intensity affects current, not kinetic energy}} \]
Step 3: Consider the remaining options. The area of the photocell and the time of illumination do not appear in Einstein's equation and hence do not determine the kinetic energy of the emitted electrons. Therefore, \[ \boxed{K_{\max}\propto \nu} \] and \[ \boxed{\text{Answer = (B)}} \]
Step 1: Observe the factors affecting kinetic energy. From \[ K_{\max}=h\nu-\phi, \] the kinetic energy depends directly on the frequency \(\nu\) of the incident photons.
Step 2: Examine the effect of intensity. Intensity controls the number of emitted photoelectrons (photoelectric current), but does not affect their maximum kinetic energy. \[ \boxed{\text{Intensity affects current, not kinetic energy}} \]
Step 3: Consider the remaining options. The area of the photocell and the time of illumination do not appear in Einstein's equation and hence do not determine the kinetic energy of the emitted electrons. Therefore, \[ \boxed{K_{\max}\propto \nu} \] and \[ \boxed{\text{Answer = (B)}} \]
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