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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
Photoelectric Effect and Wave–Particle Duality

Concepts Of Physics MCQ Edition [Volume 2] · Physics · Photoelectric Effect and Wave–Particle Duality

Assuming the sun is directly overhead, the surface of the earth...

Assuming the sun is directly overhead, the surface of the earth receives sunlight at 1.4imes103 W m−21.4 imes 10^3 \text{ W m}^{-2}. Treat the light as monochromatic with an average wavelength of 500 nm500 \text{ nm}, and assume no light is absorbed between the sun and the earth's surface. The distance from the sun to the earth is 1.5×1011 m1.5 \times 10^{11} \text{ m}. Determine the number of photons falling per second on each square metre of the earth's surface directly below the sun.

Options

  1. A3.5×10213.5 \times 10^{21}
  2. B7.0×10217.0 \times 10^{21}
  3. C3.5×10193.5 \times 10^{19}
  4. D1.4×10211.4 \times 10^{21}

Correct answer

A. 3.5×10213.5 \times 10^{21}

Step-by-step solution

Intensity of sunlight on the earth's surface, I=1.4×103 W m−2I = 1.4 \times 10^3 \text{ W m}^{-2}

Energy of a single photon is given by E=hcλE = \dfrac{hc}{\lambda}

Substituting the given values:

E=6.63×10−34×3×108500×10−9=3.98×10−19 JE = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.98 \times 10^{-19} \text{ J}

The number of photons falling per second per square metre is n=IEn = \dfrac{I}{E}

n=1.4×1033.98×10−19≈3.5×1021n = \dfrac{1.4 \times 10^3}{3.98 \times 10^{-19}} \approx 3.5 \times 10^{21}
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