The number of photons of wavelength 540 nm e | vedclass

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

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$3 	imes 10^{20}$

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(C) The power $P$ of the bulb is the total energy emitted per second,given by $P = \frac{nE}{\Delta t}$,where $n$ is the number of photons emitted per second and $E = \frac{hc}{\lambda}$ is the energy of one photon.Given: $P = 100 \ W$,$\lambda = 540 \ nm = 540 	imes 10^{-9} \ m$,$h = 6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.Substituting these values into the formula $P = \frac{nhc}{\lambda}$:$100 = \frac{n 	imes (6 	imes 10^{-34}) 	imes (3 	imes 10^8)}{540 	imes 10^{-9}}$$100 = \frac{n 	imes 18 	imes 10^{-26}}{540 	imes 10^{-9}}$$100 = n 	imes \frac{18}{540} 	imes 10^{-17}$$100 = n 	imes \frac{1}{30} 	imes 10^{-17}$$n = 100 	imes 30 	imes 10^{17} = 3000 	imes 10^{17} = 3 	imes 10^{20}$Thus,the number of photons emitted per second is $3 	imes 10^{20}$.

The number of photons of wavelength $540 \ nm$ emitted per second by an electric bulb of power $100 \ W$ is (taking $h = 6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$):

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