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(D) The power $P$ of the bulb is the total energy emitted per second. The energy of a single photon is given by $E = \frac{hc}{\lambda}$.If $n$ is the

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$2 	imes 10^{20}$

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(D) The power $P$ of the bulb is the total energy emitted per second. The energy of a single photon is given by $E = \frac{hc}{\lambda}$.If $n$ is the number of photons emitted per second,then the total power is $P = n 	imes E = n \frac{hc}{\lambda}$.Rearranging for $n$,we get $n = \frac{P \lambda}{hc}$.Given: $P = 66 \ W$,$\lambda = 600 \ nm = 600 	imes 10^{-9} \ m$,$h = 6.6 	imes 10^{-34} \ J \cdot s$,and $c = 3 	imes 10^8 \ m/s$.Substituting the values:$n = \frac{66 	imes 600 	imes 10^{-9}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}$$n = \frac{66 	imes 600 	imes 10^{-9}}{19.8 	imes 10^{-26}}$$n = \frac{39600 	imes 10^{-9}}{19.8 	imes 10^{-26}}$$n = 2000 	imes 10^{17} = 2 	imes 10^{20}$.Thus,the number of photons emitted per second is $2 	imes 10^{20}$.

The number of photons emitted per second by a bulb of $66 \ W$ power emitting waves of wavelength $600 \ nm$ is . . . . . . . $(h = 6.6 \times 10^{-34} \ J \cdot s)$

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