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Question: Light rays of wavelengths 6000 Å and of photon intensity 39.6 watt/m<sup>2</sup> is incident on a me...

Light rays of wavelengths 6000 Å and of photon intensity 39.6 watt/m2 is incident on a metal surface. If only one percent of photons incident on surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be : (h = 6.64 × 10–34 J-s, velocity of light = 3 × 108 m/s)

A

12 × 1018

B

10 × 1018

C

12 × 1017

D

12 × 1016

Answer

12 × 1017

Explanation

Solution

I =

39.6 = n×6.6×10−34×3×1081×1×6000×10−10\frac { \mathrm { n } \times 6.6 \times 10 ^ { - 34 } \times 3 \times 10 ^ { 8 } } { 1 \times 1 \times 6000 \times 10 ^ { - 10 } }