Light of particular frequency v is incident on a metal surface. When the intensity of incident radiation is increased, the photoelectric current

Decreases

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Increases

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Remains unchanged

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Sometimes increases and sometimes decreases

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Solution

The correct option is C Increases
Intensity of light $\propto$ No. of photons striking the metal per unit time
No. of photons striking the metal $\propto$ Electrons knocked out of metal per unit time.
Current = $C h a r g e / t i m e$
More the electrons knocked out per unit time, more is the photo electric current.

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Light of particular frequency v is incident on a metal surface. When the intensity of incident radiation is increased, the photoelectric current

The correct option is C IncreasesIntensity of light ∝ No. of photons striking the metal per unit time No. of photons striking the metal ∝ Electrons knocked out of metal per unit time.Current = Charge/timeMore the electrons knocked out per unit time, more is the photo electric current.