Question:
A light of wavelength $\lambda$ and intensity $I$ falls on photosensitive material. If $N$ photoelectrons are emitted, each with kinetic energy $E$, then
A light of wavelength $\lambda$ and intensity $I$ falls on photosensitive material. If $N$ photoelectrons are emitted, each with kinetic energy $E$, then
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In photoelectric effect, intensity controls the number of electrons, while wavelength controls their kinetic energy.
Updated On: Feb 4, 2026
- $E \propto \dfrac{1}{\lambda},\ N \propto I$
- $E \propto I,\ N \propto I$
- $E \propto I,\ N \propto \dfrac{1}{\lambda}$
- $E \propto I,\ N \propto \lambda$
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The Correct Option is A
Solution and Explanation
Step 1: Relation between kinetic energy and wavelength.
According to photoelectric equation: \[ E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi \] Thus, kinetic energy of photoelectrons depends on wavelength, not on intensity.
Step 2: Dependence of photoelectron number on intensity.
The number of emitted photoelectrons is directly proportional to the intensity of incident light.
Step 3: Final conclusion.
\[ E \propto \dfrac{1}{\lambda}, \quad N \propto I \]
According to photoelectric equation: \[ E = h\nu - \phi = \dfrac{hc}{\lambda} - \phi \] Thus, kinetic energy of photoelectrons depends on wavelength, not on intensity.
Step 2: Dependence of photoelectron number on intensity.
The number of emitted photoelectrons is directly proportional to the intensity of incident light.
Step 3: Final conclusion.
\[ E \propto \dfrac{1}{\lambda}, \quad N \propto I \]
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