When the sun is directly overhead, the surface of the earth receives 1.4 xx `(10^3) W (m^-2)` of sunlight. Assume that the light is monochromatic with average wavelength 500mn 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 xx `(10^11)` 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?

Explanation

(a) Here Intensity , ` I = 1.4 xx 10^(3) omega / m^(2)`,
Intensity , `J = Power / Area = 1.4 xx 10^(3) omega/ m^(2)`
Let no. of photons / sec emitted = n
` :. Power = Energy emitted / sec `
` = nhc / lambda = P `
`No. of photons / m^(2) = nhc / lambda = intensity `
`n = Intensity xx lambda / hc`
`= 1.9 xx 10^(3) xx 5 xx 10^(-9) / 6.63 xx 10^(-34) xx 3 xx 10^(8)`
` = 3.5 xx 10^(21)`
(b) Consider no. of two parts at a distance r and r + dr from the source.
The time interval `dlt` in which the phton travels from one point to another = dt In this time the total no. of photons emitted
` = N = ndt = (p lambda / gc) dr / C `
These points will be presented between two spherical shells of rail ' r ' and ' r +dr ' . It is the distance of the 1st point from the sources. In this case ` l = 1.5 xx 10^(11) m `
`lambda = 500 nm = 500 xx 10^(-9) m `
P / 4 pir r^(2) = 1.4 xx 10^(5) `
`:. No. of photons / m ^(3) = P / 4 pie r^(2) lambda / hc^(2)`
` = 1.4 xx 10^(3) xx 500 xx 10^(-9) / 6.63 xx 10^(-34) xx 3 xx 10^(8)`
`= 1.2 xx 10^(13)`
(d) No. of photons (No. of photons / sec / m^(2) `
`xx Area`
`= (3.5 xx 10^(21) ) xx 4 F `
` = 3.5 xx 10^(21) xx 4 ^(0.14) (1.5 xx 10^(11))^(2)`
`= 9.9 xx 10^(44)`