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If an X-ray tube is operating at a current of 30.0 mA:a) How many electrons are striking the target per second? b) If the potential difference between the anode and cathode of this tube is 100.0 kV, how much power is expended

User Klaasvaak
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1 Answer

5 votes

Answer:

Step-by-step explanation:

A. Q= It

30E-3A=q

Ne = q

30E-3/1.6*10-19= N

N= 1.8*10^16 electrons

B. Power= I x v

= 30*10^-3A x 100*10^3v

= 3000watts

User Piyush Patil
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