Find the number of photons emitted per second | vedclass

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(A) The current $I$ in a conductor is given by $I = n e A v_d$, where $n$ is the electron density, $e$ is the charge of an electron, $A$ is the cross-

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$6.25 	imes 10^{19}$

Vedclass · Answer

(A) The current $I$ in a conductor is given by $I = n e A v_d$, where $n$ is the electron density, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.Substituting the given values: $I = 10^{20} 	imes 1.6 	imes 10^{-19} 	imes 0.25 	imes 1.5$.$I = 16 	imes 0.25 	imes 1.5 = 4 	imes 1.5 = 6 \ A$.The number of electrons emitted per second is $N_e = I / e = 6 / (1.6 	imes 10^{-19}) = 3.75 	imes 10^{19} \ 	ext{electrons/s}$.Given that $60\%$ of emitted photons result in electron emission, let $N_p$ be the number of photons emitted per second: $0.60 	imes N_p = N_e$.$N_p = N_e / 0.60 = (3.75 	imes 10^{19}) / 0.6 = 6.25 	imes 10^{19} \ 	ext{photons/s}$.

Column-$I$	Column-$II$
$A$. Hallwachs $\&$ Lenard	$P$. Transformers
$B$. Franck-Hertz	$Q$. Microwave
$C$. Klystron valve	$R$. Quantization of energy levels
$D$. Nicola Tesla	$S$. Photoelectric effect

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Column-$I$ Column-$II$

$A$. Hallwachs $\&$ Lenard $P$. Transformers

$B$. Franck-Hertz $Q$. Microwave

$C$. Klystron valve $R$. Quantization of energy levels

$D$. Nicola Tesla $S$. Photoelectric effect

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