An electric bulb of 100 W power emits photon | vedclass

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(C) The power of the bulb is $P = 100 \ W$,which means it emits $100 \ J$ of energy per second.The energy of a single photon is given by the formula $

Vedclass · Accepted Answer

$2.06 	imes 10^{20}$

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(C) The power of the bulb is $P = 100 \ W$,which means it emits $100 \ J$ of energy per second.The energy of a single photon is given by the formula $E = \frac{hc}{\lambda}$.Given values: $h = 6.63 	imes 10^{-34} \ J \cdot s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 410 	imes 10^{-9} \ m$.Substituting these values,the energy of one photon is $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{410 	imes 10^{-9}} \approx 4.85 	imes 10^{-19} \ J$.The number of photons emitted per second $(n)$ is calculated as $n = \frac{P}{E}$.$n = \frac{100}{4.85 	imes 10^{-19}} \approx 2.06 	imes 10^{20}$ photons per second.

An electric bulb of $100 \ W$ power emits photons of wavelength $410 \ nm$ per second. The number of photons emitted per second is: $(h = 6.63 \times 10^{-34} \ J \cdot s, c = 3 \times 10^8 \ m/s)$

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