The potential difference applied to an X-ray | vedclass

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(D) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t} = \frac{ne}{t}$.Here,$n$ is the number of electrons striking th

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$4 	imes 10^{16}$

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(D) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t} = \frac{ne}{t}$.Here,$n$ is the number of electrons striking the target per second,$e$ is the charge of an electron $(1.6 	imes 10^{-19} \ C)$,and $I$ is the current $(6.4 \ mA = 6.4 	imes 10^{-3} \ A)$.Rearranging the formula to solve for $n$:$n = \frac{I}{e} = \frac{6.4 	imes 10^{-3}}{1.6 	imes 10^{-19}}$$n = 4 	imes 10^{16} \ 	ext{electrons/second}$.Thus,the correct option is $D$.

The potential difference applied to an $X$-ray tube is $5 \ KV$ and the current through it is $6.4 \ mA$. The number of electrons striking the target per second is:

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