An X-ray tube operates at 50 kV. It converts | vedclass

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(A) The total power input to the $X$-ray tube is $P_{in} = VI$, where $V = 50 \ kV = 50 	imes 10^3 \ V$ and $I$ is the current.Since $1\%$ of the ene

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$6.25 	imes 10^{16}$

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(A) The total power input to the $X$-ray tube is $P_{in} = VI$, where $V = 50 \ kV = 50 	imes 10^3 \ V$ and $I$ is the current.Since $1\%$ of the energy is converted into $X$-rays, the remaining $99\%$ is converted into heat.Given that the heat produced is $495 \ W$, we have:$P_{heat} = 0.99 	imes P_{in} = 0.99 	imes V 	imes I$$495 = 0.99 	imes (50 	imes 10^3) 	imes I$$I = \frac{495}{0.99 	imes 50 	imes 10^3} = \frac{495}{49500} = 0.01 \ A = 10 \ mA$.The number of electrons striking the target per second is given by $n = \frac{I}{e}$, where $e = 1.6 	imes 10^{-19} \ C$.$n = \frac{0.01}{1.6 	imes 10^{-19}} = \frac{10^{-2}}{1.6 	imes 10^{-19}} = 0.625 	imes 10^{17} = 6.25 	imes 10^{16}$ electrons per second.

An $X$-ray tube operates at $50 \ kV$. It converts $1\%$ of the energy into $X$-rays. If the amount of heat produced is $495 \ W$, then the number of electrons striking the target per second is:

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