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Physics Question on Photoelectric Effect

Ultraviolet light wavelength 300 nm300 \,nm and intensity 1.0 Wm−21.0 \,Wm ^{-2} falls on the surface of a photoelectric material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 cm21.0 \,cm ^{2} of the surface is nearly

A

9.61×10149.61\times {{10}^{14}}

B

4.12×10134.12\times {{10}^{13}}

C

1.51×10121.51\times {{10}^{12}}

D

2.13×10112.13\times {{10}^{11}}

Answer

1.51×10121.51\times {{10}^{12}}

Explanation

Solution

Energy of each photon,
E=hcλ=6.6×10−34×3×108300×10−9E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}}
E=6.6×10−19JE=6.6 \times 10^{-19} J
Power of source
P=IA=1.0×1.0×10−4=10−4WP=I A=1.0 \times 1.0 \times 10^{-4}=10^{-4} W
So, number of photons per sec as
Nt=PE=10−46.6×10−19\frac{N}{t}=\frac{P}{E}=\frac{10^{-4}}{6.6 \times 10^{-19}}
Number of electrons emitted is
N′=1100×10−46.6×10−19N' =\frac{1}{100} \times \frac{10^{-4}}{6.6 \times 10^{-19}}
⇒N′=1.51×1012/s\Rightarrow N '=1.51 \times 10^{12} / s