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Question

(C) The power of the bulb is $P = 200 \ W = 200 \ J \ s^{-1}$.The wavelength of the light is $\lambda = 200 \ nm = 200 	imes 10^{-9} \ m = 2 	imes 1

Vedclass · Accepted Answer

$2.012 	imes 10^{20}$

Vedclass · Answer

(C) The power of the bulb is $P = 200 \ W = 200 \ J \ s^{-1}$.The wavelength of the light is $\lambda = 200 \ nm = 200 	imes 10^{-9} \ m = 2 	imes 10^{-7} \ m$.The energy of one photon is $E = \frac{hc}{\lambda}$.Let $n$ be the number of photons emitted per second.The total energy emitted per second is $P = n 	imes E = n 	imes \frac{hc}{\lambda}$.Rearranging for $n$: $n = \frac{P 	imes \lambda}{h 	imes c}$.Substituting the values: $n = \frac{200 	imes 2 	imes 10^{-7}}{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}$.$n = \frac{400 	imes 10^{-7}}{19.878 	imes 10^{-26}} \approx 2.012 	imes 10^{20}$ photons per second.

$A$ $200 \ W$ bulb emits monochromatic light of wavelength $200 \ nm$. Calculate the number of photons emitted per second by the bulb.

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