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QPages 365-366: Q5: HC Verma: Concepts of Physics - Volume 2 Photoelectric Effect and Wave-Particle Duality Pages 365-366: Q5 Solution: When the sun is directly overhead, the surface of the earth…

⚠️ Common Exam Traps & Conceptual Pitfalls

When converting the intensity $I$ to a photon count, remember that $I$ gives the energy flux ($ ext{J}, ext{s}^{-1}, ext{m}^{-2}$). The photon flux (photons m$^{-2}$ s$^{-1}$) is $I/E$, but the photon density (photons m$^{-3}$) is that flux divided by the speed of light: $N=I/(E,c)$. Forgetting the $c$ factor leads to an overestimate of the density by three orders of magnitude.

When the sun is directly overhead, the surface of the earth receives 1.4×103 W m−21.4 \times 10^3 \text{ W m}^{-2} of sunlight. Assume that the light is monochromatic with average wavelength 500 nm and that no light is absorbed in between the sun and the earth's surface. The distance between the sun and the earth is 1.5×1011 m1.5 \times 10^{11} \text{ m}. a) Calculate the number of photons falling per second on each square metre of earth's surface directly below the sun. b) How many photons are there in each cubic metre near the earth's surface at any instant? c) How many photons does the sun emit per second? Use h=6.63×10−34 J-s=4.14×10−15 eV-sh = 6.63 \times 10^{-34} \text{ J-s} = 4.14 \times 10^{-15} \text{ eV-s}, c=3×108 m/sc = 3 \times 10^8 \text{ m/s} and me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \text{ kg})

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**Answer:** (a) n=3.5×1021n = 3.5 \times 10^{21} photons/m_/s (b) N=5.9×106N = 5.9 \times 10^6 photons/m_ (c) nsun=9.9×1044n_{sun} = 9.9 \times 10^{44} photons/s. **Step-by-Step Explanation:** Energy per photon: E=1240500=2.48E = \frac{1240}{500} = 2.48 eV =3.97×10−19= 3.97 \times 10^{-19} J. (a) Photons per m_ per second: n=1.4×1033.97×10−19=3.5×1021n = \frac{1.4 \times 10^3}{3.97 \times 10^{-19}} = 3.5 \times 10^{21}. (b) Photon density: N=IEc=1.4×1033.97×10−19×3×108=5.9×106N = \frac{I}{Ec} = \frac{1.4 \times 10^3}{3.97 \times 10^{-19} \times 3 \times 10^8} = 5.9 \times 10^6 photons/m_. (c) Total photons from sun: nsun=3.5×1021×4π(1.5×1011)2=9.9×1044n_{sun} = 3.5 \times 10^{21} \times 4\pi (1.5 \times 10^{11})^2 = 9.9 \times 10^{44} photons/s.

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