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(C) The power of the bulb is $P = 60 \ W$,which means $60 \ J$ of energy is emitted per second.Let $n$ be the number of photons emitted per second.The

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$2 	imes 10^{20}$

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(C) The power of the bulb is $P = 60 \ W$,which means $60 \ J$ of energy is emitted per second.Let $n$ be the number of photons emitted per second.The energy of one photon is given by $E_{photon} = \frac{hc}{\lambda}$.The total energy emitted per second is $E = n 	imes E_{photon} = n \frac{hc}{\lambda}$.Rearranging for $n$,we get $n = \frac{E \lambda}{hc}$.Substituting the given values: $E = 60 \ J$,$\lambda = 660 	imes 10^{-9} \ m$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.$n = \frac{60 	imes 660 	imes 10^{-9}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{39600 	imes 10^{-9}}{19.8 	imes 10^{-26}} = 2000 	imes 10^{17} = 2 	imes 10^{20}$ photons/second.

The number of photons emitted per second from a $60 \ W$ light bulb is .......... . The wavelength of the photon is $660 \ nm$. $(h = 6.6 \times 10^{-34} \ J \cdot s)$

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