Question

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.

Answer 1

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

Answer 2

thks ujjwal

Answer 3

WC :)

Answer 4

what does n stand for?

Answer 5

correct please give best response

Answer 6

n is number of photons!