A 200, W sodium street lamp emits yellow ligh | vedclass

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(B) The power of the lamp is $P = 200\, W$. Since it is $50\%$ efficient,the power converted into light is $P_{	ext{light}} = 0.50 	imes 200\, W = 1

Vedclass · Accepted Answer

$3 	imes 10^{20}$

Vedclass · Answer

(B) The power of the lamp is $P = 200\, W$. Since it is $50\%$ efficient,the power converted into light is $P_{	ext{light}} = 0.50 	imes 200\, W = 100\, W$.The energy of a single photon is given by $E = \frac{hc}{\lambda}$,where $h = 6.63 	imes 10^{-34}\, J\cdot s$,$c = 3 	imes 10^8\, m/s$,and $\lambda = 0.6 	imes 10^{-6}\, m$.Let $n$ be the number of photons emitted per second. The total power is $P_{	ext{light}} = nE = n \frac{hc}{\lambda}$.Rearranging for $n$: $n = \frac{P_{	ext{light}} \lambda}{hc} = \frac{100 	imes 0.6 	imes 10^{-6}}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}$.Using the approximation $hc \approx 20 	imes 10^{-26}\, J\cdot m$ (or $1.989 	imes 10^{-25}\, J\cdot m$):$n = \frac{60 	imes 10^{-6}}{19.89 	imes 10^{-26}} \approx 3.01 	imes 10^{20}$.Thus,the number of photons emitted per second is $3 	imes 10^{20}$.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

$A$ $200\, W$ sodium street lamp emits yellow light of wavelength $0.6\, \mu m$. Assuming it to be $50\%$ efficient in converting electrical energy to light,the number of photons of yellow light it emits per second is:

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Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

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