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# The energy flux of sunlight reaching the surface of the earth is $1.388 \times 10^3 W/m^2$. 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 $550\,nm$.

Last updated date: 15th Aug 2024
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Hint: The energy flux of sunlight is the power of sunlight reaching the surface of the earth per ${m^2}$.
Power can also be written as the product of energy of each photon and the total number of photons.$P = nE$
Energy of photon,$E$ is given by the equation
$E = h\upsilon$
Where, $\upsilon$ is the frequency and $h$ is the Planck’s constant.
The value of Planck’s constant is
$h = 6.626 \times {10^{ - 34}}\,{m^2}kg{s^{ - 1}}$
We know $c = \upsilon \lambda$
Where, $c$ is the speed of light and $\lambda$ is the wavelength.
Therefore,
$\upsilon = \dfrac{c}{\lambda }$

Complete step by step solution:
The energy flux of sunlight or the power of sunlight reaching the surface of the earth per ${m^2}$ is $\phi = 1.388 \times {10^3} W/{{m^2}}$.
That is power, $P = 1.388 \times {10^3}\,W$
Average wavelength,
$\lambda = 550\,nm \\ = 550 \times {10^{ - 9}}\,m \\$.
We need to find the number of photons per square meter that are incident on the Earth per second.
Let this number be $n$ .
Power can also be written as the product of energy of each photon and the total number of photons.$P = nE$
Energy of photon,$E$ is given by the equation
$E = h\upsilon$...........….. (1)
Where, $\upsilon$ is the frequency and $h$ is the Planck’s constant.
The value of Planck’s constant is
$h = 6.626 \times {10^{ - 34}}\,{m^2}kg{s^{ - 1}}$
In the question the wavelength of the photon is given. Hence, we need to write equation (1) in terms of wavelength.
We know $c = \upsilon \lambda$
Where, $c$ is the speed of light.
Therefore,
$\upsilon = \dfrac{c}{\lambda }$..........……(2)
Substitute equation (2) in equation (1). We get
$E = h\dfrac{c}{\lambda }$
Therefore,
$P = nE \\ = nh\dfrac{c}{\lambda } \\$
We need to find $n$
That is,
$n = \dfrac{{P\lambda }}{{hc}}$
Now, substitute the given values.
$n = \dfrac{{1.388 \times {{10}^3}\, \times 550 \times {{10}^{ - 9}}\,}}{{6.626 \times {{10}^{ - 34}}\, \times 3 \times {{10}^8}}} \\ = 3.847 \times {10^{21}}\, \\$
So, the number of photons per square meter that are incident on the Earth per second is $3.847 \times {10^{21}}\,$.

Note: Formulae to remember-
$P = nE$
Where, $n$ is the number of photons and $E$ is the energy of each photon.
$E = h\upsilon$
Where, $\upsilon$ is the frequency and $h$ is the Planck’s constant
$c = \upsilon \lambda$
Where, $c$ is the speed of light, $\upsilon$ is the frequency and $\lambda$ is the wavelength.