The beam of light has three wavelengths 4144 | vedclass

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Question

(b)

$\left(\lambda_0\right)=\frac{ hc }{\phi}=5404\,A$

For each wavelength energy incident on the surface per unit time

= intensity of each $

Vedclass · Accepted Answer

$1.075 	imes 10^{12}$

Vedclass · Answer

(b)

$\left(\lambda_0ight)=\frac{ hc }{\phi}=5404\,A$

For each wavelength energy incident on the surface per unit time

= intensity of each $	imes$ area of the surface wavelength $=1.2 	imes 10^{-7}$ joule

$E=\left(1.2 	imes 10^{-7}ight) 	imes 2=2.4 	imes 10^{-7}\,J$

Number of photons $n _1$ due to wavelength $4144\,A$

$n_1=\frac{\left(2.4 	imes 10^{-7}ight)\left(4144 	imes 10^{-10}ight)}{\left(6.63 	imes 10^{-34}ight)\left(3 	imes 10^8ight)}=0.5 	imes 10^{12}$

Number of photons $n_2$ due to the wavelength $4972 \mathring  A$

$n _2=\frac{\left(2.4 	imes 10^{-7}ight)\left(4972 	imes 10^{-10}ight)}{\left(6.63 	imes 10^{-34}ight)\left(3 	imes 10^3ight)}=0.572 	imes 10^{12} $

$N = n _1+ n _2=1.075 	imes 10^{12}$

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