Question

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Q- the energy flux of Sunlight reaching the surface of the Earth is 1.39 into 10 to the power 3 W m ^-2. if the average wavelength of Sunlight is 550 nm then what will be the number of photons per square metre incident on the earth .​

Answer 1

Explanation:

Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 103 W/m2

Hence, power of sunlight per square metre, P = 1.388 × 103 W

Speed of light, c = 3 × 108 m/s

Planck’s constant, h = 6.626 × 10−34 Js

Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10-9 m

Number of photons per square metre incident on earth per second = n Hence, the equation for power can be written as: P= nE

∴ n = P/E = Pλ / hc = 1.388 x 103 x 550 x 10-9 / 6.626 x 10-34 x 3 x 108 = 3.84 x 1021 photons/m2/s

Therefore, every second, 3.84 x 1021 photons are incident per square metre on earth.

Answer 2

Index :-

$N_P=Number\ of\ incident\ photons$

$h=Plancks\ Constant$

$c=Speed\ of\ light$

${\lambda}=Incident\ wavelength$

$T=Time$

$A=Area$

$I=Intensity\ of\ sunglight\ reaching\ on\ the\ earth\ surface$

Solution :-

Energy flux is equivalent to intensity.

Intensity of sunlight reaching the earth surface (I) is

$=\frac{(Number\ of\ photons\ emitted)(Energy\ of\ 1\ photon)}{(Time)(Area)}$$=\frac{(Number\ of\ photons\ emitted)(\frac{hc}{ \lambda }) }{(Time)(Area)}$

$=\frac{(N_P)(\frac{hc}{\lambda}) }{(T)(A)}$

$=>N_P=\frac{(I)(T)(A)(\lambda)}{hc}$

$=>N_P=\frac{((1.39\times10^3)W/m^2)((1)s)((1)m^2)((550)nm)}{((1240)eV)}$

$=>N_P=3.85\times10^{21}$

Hence, number of photons incident on the Earth per second/square meter are 3.85 × 10²¹.

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