In X-ray tube the potential difference between the anode and the cathode is 20 kV and the current flowing is 1.6 mA. The number of electrons striking the anode in 1s is ..... (Take `e=1.6xx10^(-19)C)`

Options

A. `10^(14)`
B. `1.25xx10^(16)`
C. `10^(16)`
D. `6.25xx10^(18)`

Answer

C

Explanation

The number of electron striking the anode in 1s,
`n=(1)/(3)=(1.6xx10^(-3))/(1.6xx10^(-19))=10^(16)`