14 times 10^{15} electrons reach the anode pe | vedclass

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(B) The current $i$ is given by the rate of flow of charge: $i = n 	imes e$,where $n = 14 	imes 10^{15} 	ext{ electrons/s}$ and $e = 1.6 	imes 10^

Vedclass · Accepted Answer

$200$

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(B) The current $i$ is given by the rate of flow of charge: $i = n 	imes e$,where $n = 14 	imes 10^{15} 	ext{ electrons/s}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$.$i = (14 	imes 10^{15}) 	imes (1.6 	imes 10^{-19}) = 22.4 	imes 10^{-4} 	ext{ A} = 2.24 	imes 10^{-3} 	ext{ A}$.The power consumed is $P = 448 	ext{ mW} = 448 	imes 10^{-3} 	ext{ W}$.Using the formula $P = V 	imes i$,we find the voltage $V = \frac{P}{i}$.$V = \frac{448 	imes 10^{-3}}{2.24 	imes 10^{-3}} = \frac{448}{2.24} = 200 	ext{ V}$.

$14 \times 10^{15}$ electrons reach the anode per second. If the power consumed is $448 \text{ mW}$,then the plate (anode) voltage is .... $V$.

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