When the sun is directly overhead, the surface of the earth receives 1.4 × 103 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 is 1.5 × 1011 m.

(a) Calculate the number of photons falling per second on each square meter of earth’s surface directly below the sun.


(b) How many photons are there is each cubic meter near the earth’s surface at any instant?


(c) How many photons does the sun emit per second?


(a); intensity at the earth surface is 1.4 × 103 W m–2


Energy of one photon is


Then the no of photon (n) required to produce J per square and per unit sec is



(b); as we have to calculate the photons in a cubic meter.


We above calculated the no photons that are falls in unit square per sec.


In a unit sec all the photons which are at height [ ] fall in unit. The photons which fall from 1 m height will fall in .


If there are photons which fall in 1 sec,


Then, no. of photons which fall in is =


These photons are in the 1 near the earth surface.


(c); As explained above, the no. of photons falling in 1 m square area per sec on earth surface is . The no of photons which fall on the surface of sphere of radius (distance between the sun and the earth surface) is the total no of photons which will be emitted by sun in 1 sec.


By this, surface area of sphere is 4π m square. Therefore, the no of photons will be




1