The intensity of light energy hitting a surface is the number of joules hitting

Question

The intensity of light energy hitting a surface is the number of joules hitting a square meter of the surface in one second. The energy density of the wave moves along with the wave. In the case of light, half the energy is carried by the E-field, and half is carried by the magnetic field. The energy density total (electric + magnetic) moves along with the wave at the speed of light c. Therefore in one second, 3 x 10^8 m lenght of wave strikes the surface. The maximum solar energy reaching the Earth's surface at noon in the summer in Arizona is about 1030 W/m^2 If this energy is concentrated in a single electromagnetic light wave, what is the electric field in the wave? Use the intensity above together with the formula relating E-field and intensity to find the E-field of the sunlight: Intensity = E^2/8 pi k *c E =_______v/m squareroot intensity/C middot 8 pi k = E squareroot 1030/3.108 middot 8 pi middot 9 middot 10^9 = Calculate the energy density arriving in each cubic meter of light arriving at Earth: E^2/8 pi k Energy density =________J/m^3 1/2 middot 8.8 small element of^-12 middot 881^1 + (1/2 middot 881^2/4 pi times 10^-7 3.09 = 10^11 How many photons hit a square meter of earth in 1 second, assuming the light is all green light with wavelength 550 nm?

Explanation / Answer

from given formula

E^2=intensity*(8pi*K)/c

E=sqrt(intensity*(8pi*K)/c)=881.25 V/m

(b) energy density=E^2/8piK=3.43*10^(-6) J/m^3

(c)energy in one square meter in 1 sec =intensity=1030 J

energy=n*h*(c/wavelength)=1030

n is number of photon ,h is plank constant ,c is speed of light

after solving we got

n=2.85*10^21

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