Find the number of photons emitted per second | vedclass

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(C) The power $P$ of the bulb is given by the total energy emitted per second,which is $P = \frac{nE}{\Delta t}$,where $n$ is the number of photons pe

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$3 	imes 10^{20}$

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(C) The power $P$ of the bulb is given by the total energy emitted per second,which is $P = \frac{nE}{\Delta t}$,where $n$ is the number of photons per second and $E = \frac{hc}{\lambda}$ is the energy of one photon.Given:Power $P = 100 \ W = 100 \ J/s$Wavelength $\lambda = 540 \ nm = 540 	imes 10^{-9} \ m$Planck's constant $h = 6 	imes 10^{-34} \ J \cdot s$Speed of light $c = 3 	imes 10^8 \ m/s$Substituting these values into the formula $P = \frac{n \cdot hc}{\lambda \cdot t}$:$100 = \frac{n 	imes (6 	imes 10^{-34}) 	imes (3 	imes 10^8)}{540 	imes 10^{-9}}$Solving for $n$:$n = \frac{100 	imes 540 	imes 10^{-9}}{6 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{54000 	imes 10^{-9}}{18 	imes 10^{-26}}$$n = 3000 	imes 10^{17} = 3 	imes 10^{20}$Thus,the number of photons emitted per second is $3 	imes 10^{20}$.

Find the number of photons emitted per second by a $100 \ W$ bulb emitting light of wavelength $540 \ nm$. (Given: $h = 6 \times 10^{-34} \ J \cdot s$)

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