The potential difference applied to an X-ray | vedclass

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(A) The current $i$ is defined as the rate of flow of charge,given by the formula $i = \frac{Ne}{t}$,where $N$ is the number of electrons,$e$ is the e

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$2 	imes 10^{16}$

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(A) The current $i$ is defined as the rate of flow of charge,given by the formula $i = \frac{Ne}{t}$,where $N$ is the number of electrons,$e$ is the elementary charge $(1.6 	imes 10^{-19} 	ext{ C})$,and $t$ is the time.To find the number of electrons striking the target per second,we rearrange the formula to solve for $\frac{N}{t}$:$\frac{N}{t} = \frac{i}{e}$Given $i = 3.2 	ext{ mA} = 3.2 	imes 10^{-3} 	ext{ A}$ and $e = 1.6 	imes 10^{-19} 	ext{ C}$:$\frac{N}{t} = \frac{3.2 	imes 10^{-3}}{1.6 	imes 10^{-19}}$$\frac{N}{t} = 2 	imes 10^{16} 	ext{ electrons/sec}$.

The potential difference applied to an $X$-ray tube is $5 \text{ kV}$ and the current through it is $3.2 \text{ mA}$. Then the number of electrons striking the target per second is

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