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`        When the sun is directly overhead, the surface of the earth receives 1.4 × 10^3 W/m^2 of sunlight. Assume that 500 nm and that no light if absorbed in between the sun and the earth’s surface. The distance between the sun and the earth’s is 1.5 × 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?`
5 years ago Navjyot Kalra
654 Points
```							Sol. a) Here intensity = l = 1.4 * 10^3 ω/m^2 Intensity, l = power/area = 1.4 * 10^3 ω/m^2
Let no. of photons/sec emitted = n ∴ Power = Energy emitted/sec = nhc/λ = P
N = intensity * λ/hc = 1.9 *10^3 *5 *10^-9/6.63 * 10^-34 *3 *10^8 = 3.5 * 10^21
b) Consider no.of two parts at a distance r and r + dr from the source.
The time interval ‘dt’ in which the photon travel from one point to another = dv/e = dt.
In this time the total no.of photons emitted = N = n dt = (pλ/hc) dr/C
These points will be present between two spherical shells of radii ‘r’ and r+dr. it is the distance of the 1st point from the sources. No. of photons per volume in the shell
(r + r + dr) = N/2πedr = Pλdr/hc^2 = 1/4π^2ch = Pλ/4πhc^2r^2
In the case = 1.5 * 10^11 m, λ = 500 nm, = 500 * 10^-9 m
P/4πr^2 = 1.4 * 10^3,  ∴ No.of photons/m^3 = P/4πr^2  λ/hc^2
c) No.of photons = (No.of photons/sec/m^2) *  Area
=  (3.5 * 10^21) * 4πr^2
3.5 * 10^21 * 4(3.14)(1.5 * 10^11)^2 * 9.9 * 10^44

```
5 years ago
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