Question

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's surface. The distance between the sun and the earth is 1.5 × 10¹¹ 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?

Answer 1

Here,
Intensity of light, I = 1.4 × 10³ W/m²,
Wavelength of light, $\lambda$ = 500 nm = 500$\times$10^$-9$ m
Distance between the Sun and Earth, l = 1.5$\times$10¹¹ m
Intensity,
$I=\frac{\mathrm{Power}}{\mathrm{Area}}=1.4\times {10}^3\mathrm{W}/{\mathrm{m}}^2$
Let n be the number of photons emitted per second.
∴ Power, P = Energy emitted/second
$P=\frac{nhc}{\lambda }$,
where $\lambda$ = wavelength of light
h = Planck's constant
c = speed of light
Number of photons/m² = $\frac{nhc}{\lambda \times A}=\frac{nhc}{\lambda \times 1}$ = I
$$\begin{gathered} \therefore n=\frac{I\times \lambda }{hc} \\ =\frac{1.4\times {10}^3\times 500\times {10}^{-9}}{6.63\times {10}^{-34}\times 3\times {10}^8} \\ =3.5\times {10}^{21} \end{gathered}$$

(b) Consider number of two parts at a distance r and r + dr from the source.
Let dt' be the time interval in which the photon travels from one part to another.
Total number of photons emitted in this time interval,
$N=ndt=\left(\frac{P\lambda }{hc\times A}\right)\frac{dr}{\mathrm{c}}$
These points will be between two spherical shells of radius 'r' and r + dr. It will be the distance of the 1st point from the sources.
In this case,
$$\begin{gathered} l=1.5\times {10}^{11}\mathrm{m} \\ \mathrm{Wavelength},\lambda =500\mathrm{nm}=500\times {10}^{-9}\mathrm{m} \\ \frac{P}{4\mathrm{\pi }{r}^2}=1.4\times {10}^3 \\ \therefore \mathrm{No}.\mathrm{of}\mathrm{photons}/{\mathrm{m}}^3=\frac{P}{4\mathrm{\pi }{r}^2}\frac{\lambda }{h{c}^2} \\ =1.4\times {10}^3\times \frac{500\times {10}^{-9}}{6.63\times {10}^{-34}\times 9\times {10}^{16}} \\ =1.2\times {10}^{13} \end{gathered}$$

(c) Number of photons emitted = (Number of photons / s-m²) × Area
$$\begin{gathered} =(3.5\times {10}^{21})\times 4\pi l^2 \\ =3.5\times {10}^{21}\times 4\times (3.14)\times (1.5\times {10}^{11}{)}^2 \\ =9.9\times {10}^{44} \end{gathered}$$