Question

The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Answer 1

Hey Dear,

◆ Answer -

3.838×10^21 photons/m^2/s

● Explaination -

# Given -

Φ = 1.388×10^3 W/m^2

λ = 550 nm = 5.5×10^-7 m

# Solution -

Number of photons per square metre area is calculated as -

n = (P/A) / E

n = Φ.λ / h.c

n = (1.388×10^3 × 5.5×10^-7) / (6.63×10^-34 × 3×10^8)

n = 3.838×10^21 photons/m^2/s

Therefore, 3.838×10^21 photons/m^2 are incident on the Earth per second.

Hope this helps

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)}$

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

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

$=>N_P=\frac{((1.388\times10^3)W/m^2)((1)s)((1)m^2)((550)nm)}{((1240)eV.nm)}$

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

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

Remarks :-

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