A blue lamp emits light of mean wavelength 45 | vedclass

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(A) Given: Wavelength $\lambda = 4500 \ 	ext{Å} = 4500 	imes 10^{-10} \ 	ext{m}$,Power $P = 150 \ 	ext{W}$,Efficiency $\eta = 8 \% = 0.08$.The pow

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$27.17 	imes 10^{18}$

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(A) Given: Wavelength $\lambda = 4500 \ 	ext{Å} = 4500 	imes 10^{-10} \ 	ext{m}$,Power $P = 150 \ 	ext{W}$,Efficiency $\eta = 8 \% = 0.08$.The power of the light emitted by the lamp is $P_{	ext{out}} = P 	imes \eta = 150 	imes 0.08 = 12 \ 	ext{W}$.The energy of a single photon is $E = \frac{hc}{\lambda}$.The number of photons emitted per second $n$ is given by $n = \frac{P_{	ext{out}}}{E} = \frac{P_{	ext{out}} 	imes \lambda}{hc}$.Using $hc \approx 1240 \ 	ext{eV} \cdot 	ext{nm} = 12400 \ 	ext{eV} \cdot 	ext{Å}$ and $1 \ 	ext{eV} = 1.6 	imes 10^{-19} \ 	ext{J}$,we have $hc = 12400 	imes 1.6 	imes 10^{-19} \ 	ext{J} \cdot 	ext{Å} = 1.984 	imes 10^{-15} \ 	ext{J} \cdot 	ext{m}$.Substituting the values: $n = \frac{12 	imes 4500 	imes 10^{-10}}{6.626 	imes 10^{-34} 	imes 3 	imes 10^8} \approx 2.717 	imes 10^{19} 	ext{ (or using } hc \approx 1.987 	imes 10^{-16} 	ext{ J} \cdot 	ext{m, } n \approx 2.717 	imes 10^{19} 	ext{)}$.Re-evaluating with standard constants: $n = \frac{12 	imes 4500 	imes 10^{-10}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8} = 2.715 	imes 10^{19}$. Given the options,the calculation $27.17 	imes 10^{18}$ is the intended answer.

$A$ blue lamp emits light of mean wavelength $4500 \ \text{Å}$. The lamp is rated at $150 \ \text{W}$ and $8 \%$ efficiency. Calculate the number of photons emitted by the lamp per second.

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