In an X-ray tube, the voltage is 5 , kV and t | vedclass

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(A) The electric current $i$ is defined as the rate of flow of charge, $i = \frac{q}{t}$.Since $q = ne$, where $n$ is the number of electrons and $e$

Vedclass · Accepted Answer

$2 	imes 10^{16}$

Vedclass · Answer

(A) The electric current $i$ is defined as the rate of flow of charge, $i = \frac{q}{t}$.Since $q = ne$, where $n$ is the number of electrons and $e$ is the elementary charge $(1.6 	imes 10^{-19} \, C)$, we have $i = \frac{ne}{t}$.To find the number of electrons per second $(n/t)$, we rearrange the formula: $\frac{n}{t} = \frac{i}{e}$.Given $i = 3.2 \, mA = 3.2 	imes 10^{-3} \, A$ and $e = 1.6 	imes 10^{-19} \, C$.Substituting the values: $\frac{n}{t} = \frac{3.2 	imes 10^{-3}}{1.6 	imes 10^{-19}} = 2 	imes 10^{16} \, 	ext{electrons/second}$.

In an $X-ray$ tube, the voltage is $5 \, kV$ and the current is $3.2 \, mA$. The number of electrons striking the anode per second is:

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