Intensity of sunlight at given position is defined as power received per unit area
so here we can say
[tex]I = 2 kJ/s*m^2[/tex]
area on which photons are received is given as
[tex]A = 4.80 cm^2 = 4.80 * 10^-4 m^2[/tex]
now we can find the power received due to sunlight
[tex]P = I*A[/tex]
[tex]P = 2* 10^3 * 4.80 * 10^-4[/tex]
[tex]P = 0.96 Watt[/tex]
now we can say this power is due to photons that strikes on surface of earth
so here we can say
[tex]P = N\frac{hc}{\lambda}[/tex]
given here that
[tex]\lambda = 510 nm[/tex]
[tex]0.96 = N\frac{6.6 * 10^{-34}* 3 * 10^8}{510*10^{-9}}[/tex]
[tex]0.96 = N * 3.88 * 10^{-19}[/tex]
[tex]N = \frac{0.96}{3.88*10^{-19}}[/tex]
[tex]N = 2.47 * 10^{18}[/tex]
so it will strike 2.47 * 10^18 photons on given area per second
