A 150 W bulb emits light of wavelength 6600 | vedclass

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(A) The power of the bulb is $P = 150 \ W$. Only $8 \%$ of the energy is emitted as light,so the power emitted as light is $P_{light} = 150 	imes \fr

Vedclass · Accepted Answer

$4 	imes 10^{19}$

Vedclass · Answer

(A) The power of the bulb is $P = 150 \ W$. Only $8 \%$ of the energy is emitted as light,so the power emitted as light is $P_{light} = 150 	imes \frac{8}{100} = 12 \ W$. The energy of one photon is given by $E = \frac{hc}{\lambda}$. The number of photons emitted per second $(n)$ is given by $n = \frac{P_{light}}{E} = \frac{P_{light} 	imes \lambda}{hc}$. Substituting the values: $n = \frac{12 	imes 6600 	imes 10^{-10}}{6.626 	imes 10^{-34} 	imes 3 	imes 10^{8}}$. Using $h = 6.6 	imes 10^{-34} \ J \cdot s$ for simplicity as per the provided solution: $n = \frac{12 	imes 6600 	imes 10^{-10}}{6.6 	imes 10^{-34} 	imes 3 	imes 10^{8}} = \frac{12 	imes 6600 	imes 10^{-10}}{19.8 	imes 10^{-26}} = 4 	imes 10^{19}$ photons per second.

$A$ $150 \ W$ bulb emits light of wavelength $6600 \ \mathring{A}$ and only $8 \%$ of the energy is emitted as light. How many light photons are emitted by the bulb per second?

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