Question

Intensity of sunlight falling normally on the earth surface is $1.4 \times 10^{3} {W m}^{- 2}$ . Assume that the light is monochromatic with average wavelength $5000 \overset{\circ}{A}$ and that no light is absorbed in between the sun and the earth’s surface. The distance between the sun and the earth is $1.5 \times 10^{11} m$ .

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

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

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

Accepted Answer

Here

Intensity of light, $I = 1.4 \times 10^{3} W m^{- 2}$

Wavelength of light, $λ = 5000 \overset{\circ}{A} = 500 \times 10^{- 9} m$

Distance between the sun and earth,

$L = 1.5 \times 10^{11} m$

Intensity, $I = \frac{power}{area} = 1.4 \times 10^{3} {W m}^{- 2}$

Let $n$ be the number of photons emitted per second,

Power $= \frac{Energy emitted}{second}$

$\Rightarrow P = \frac{n h c}{λ}$

where $λ =$ wavelength of light

$h =$ Plank's constant

$c =$ speed of light

$\frac{Number of photons}{m^{2}} = I n t e n s i t y \Rightarrow \frac{n h c}{λ \times 1} = I$

$(a)$

$= \frac{1.4 \times 10^{3} \times 500 \times 10^{- 9}}{6.63 \times 10^{- 34} \times 3 \times 10^{8}} = 3.5 \times 10^{21}$

$(b)$ Consider number of two parts at a distance $r$ and $r + d r$ from the source.

Let $d t$ be the time interval in which the photon travels from one part to another.

Total number of photons emitted in this time interval,

$N = n d t = (\frac{P λ}{h c \times A}) \frac{d r}{c}$

These points will be between two spherical shells of radius $r and r + d r$ . It will be the distance of the first point from the sources.

In this case,

$L = 1.5 \times 0^{11} m$

Wavelength, $λ = 5000 \overset{\circ}{A} = 500 \times 10^{- 9} m$

$\frac{P}{4 π r^{2}} = 1.4 \times 10^{3}$

$(c)$ Number of photons emitted $=$ number of photons $\times$ area

$= (3.5 \times 10^{21}) \times 4 π L^{2}$

$= 3.5 \times 10^{21} \times 4 \times 3.14 \times {(1.5 \times 10^{11})}^{2} = 9.9 \times 10^{44}$

Important Questions on Dual Nature of Matter and Radiation

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