Ultraviolet light of wavelength 300 nm and intensity 1.0 watt/m2 falls on the surface of a photosensitive material. lf 1% of the incident photons produce photoelectrons,then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly

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9.61×1014 per sec

4.12×1013 per sec

1.51×1012 per sec

2.13×1011 per sec

answer is C.

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Detailed Solution

Intensity of light,I= Power Area =nhcAλ⇒ Number of photon per sec, n=IAλhc∴ Number of photo electron per sec =1100×IAλhc=1100×1×10−4×300×10−96.6×10−34×3×108=1.51×1012

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