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rva.raghav

  • 2 years ago

At noon on a sunny day, there is a solar irradiance of 1 kW/m2 impinging on a solar panel that is 1 meter by 1.6 meters. To simplify the problem we assume here that the light is all the same wavelength λ=732 nm.Determine the number of photons per second incident on the panel.

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  1. ujjwal
    • 2 years ago
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    The area of solar panel is 1.6 \(m^2\). The energy of photons striking 1 \(m^2\) per sec is 1000 J So, the energy of photons striking 1.6 \(m^2\) per sec is 1.6 \(\times\)1000=1600 J Now you need to find the number of photons which will have energy of 1600 J let that number be n So,\[nh\frac{c}{\lambda}=1600\]where h is plank's constant and c is velocity of light you have \(\lambda\). Find n

  2. rva.raghav
    • 2 years ago
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    thks ujjwal

  3. ujjwal
    • 2 years ago
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    WC :)

  4. rva.raghav
    • 2 years ago
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    what does n stand for?

  5. rva.raghav
    • 2 years ago
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    correct please give best response

  6. ujjwal
    • 2 years ago
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    n is number of photons!

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