A bulb emits electromagnetic radiation of 660 | vedclass

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(D) The energy of a single photon is given by $E_{photon} = \frac{hc}{\lambda}$.The total energy $E$ is given by $E = n 	imes E_{photon} = \frac{nhc}

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$10$

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(D) The energy of a single photon is given by $E_{photon} = \frac{hc}{\lambda}$.The total energy $E$ is given by $E = n 	imes E_{photon} = \frac{nhc}{\lambda}$,where $n$ is the number of photons.Given:$E = 3 	imes 10^{-18} \, J$$\lambda = 660 \, nm = 660 	imes 10^{-9} \, m$$h = 6.6 	imes 10^{-34} \, J \cdot s$$c = 3 	imes 10^{8} \, m/s$Rearranging for $n$:$n = \frac{E 	imes \lambda}{h 	imes c}$Substituting the values:$n = \frac{3 	imes 10^{-18} 	imes 660 	imes 10^{-9}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^{8}}$$n = \frac{3 	imes 660 	imes 10^{-27}}{6.6 	imes 3 	imes 10^{-26}}$$n = \frac{1980 	imes 10^{-27}}{19.8 	imes 10^{-26}} = 100 	imes 10^{-1} = 10$Thus,the number of emitted photons is $10$.

$A$ bulb emits electromagnetic radiation of $660 \, nm$ wavelength. The total energy of radiation is $3 \times 10^{-18} \, J$. The number of emitted photons will be :
$(h = 6.6 \times 10^{-34} \, J \cdot s, \, c = 3 \times 10^{8} \, m/s)$

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$A$ bulb emits electromagnetic radiation of $660 \, nm$ wavelength. The total energy of radiation is $3 \times 10^{-18} \, J$. The number of emitted photons will be :$(h = 6.6 \times 10^{-34} \, J \cdot s, \, c = 3 \times 10^{8} \, m/s)$