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The potential difference applied to an X...

The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then the number of electros striking the target par second is

A

`1xx10^17`

B

`4xx10^15`

C

`5xx10^16`

D

`2xx10^16`

Text Solution

Verified by Experts

The correct Answer is:
D
`i= q/t = (Ne)/(t)`
`therefore N/t = i/e = (3.2xx10^(-3))/(1.6xx10^(-19)) = 2xx10^16 //sec`
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