Suppose that an x-ray tube, operating for 25 ms at 125 kvp has generated 2.5 × 1014 photons off the anode and through the filter. During this time, what was the magnitude of the tube current?

Respuesta :

given that tube is operated for t = 25 ms

which means the time for which it is used is t = 0.025 s

now the photons generated is given as

[tex]N = 2.5 * 10^{14}[/tex]

each photon will produce 1 electron as we can assume it to be 100% efficient

so number of electrons per second will be same as number of photons per second

now in order to find the current we can say

current = rate of flow of charge

[tex]i = \frac{dN}{dt}*e[/tex]

[tex]i = \frac{2.5 * 10^{14}}{0.025}* 1.6 * 10^{-19}[/tex]

[tex]i = 1.6 * 10^{-3} A[/tex]

so the current in the tube for given time will be 1.6 mA