A cathode emits 1.8 times 10^{14} electrons p | vedclass

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(B) The current $i$ is defined as the rate of flow of charge,given by $i = \frac{q}{t} = \frac{ne}{t}$.Given,the number of electrons emitted per secon

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$29$

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(B) The current $i$ is defined as the rate of flow of charge,given by $i = \frac{q}{t} = \frac{ne}{t}$.Given,the number of electrons emitted per second $\frac{n}{t} = 1.8 	imes 10^{14} \ s^{-1}$.The charge of an electron $e = 1.6 	imes 10^{-19} \ C$.Substituting the values:$i = (1.8 	imes 10^{14}) 	imes (1.6 	imes 10^{-19}) \ A$$i = 2.88 	imes 10^{-5} \ A$$i = 28.8 	imes 10^{-6} \ A$Since $1 \ \mu A = 10^{-6} \ A$,we have $i = 28.8 \ \mu A$.Rounding to the nearest integer,the maximum anode current is approximately $29 \ \mu A$.

$A$ cathode emits $1.8 \times 10^{14}$ electrons per second when heated. When $400 \ V$ is applied to the anode,all the emitted electrons reach the anode. The charge on an electron is $1.6 \times 10^{-19} \ C$. The maximum anode current is ............ $\mu A$.

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