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In an 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 1 s is ______.

Asked in GUJCET 2019 · Current and charge flow

• (1) 10¹⁴
• (2) 1.25×10¹⁶
• (3) 10¹⁶
• (4) 6.25×10¹⁸

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 10¹⁴. Sanity check: I/e = (1.6×10⁻³)/(1.6×10⁻¹⁹) = 10¹⁶ per second.
• (2) No single slip gives 1.25×10¹⁶; it would need a current of 2 mA. Sanity check: with I = 1.6 mA the 1.6 cancels exactly, leaving 10¹⁶.
• (4) 6.25×10¹⁸ is 1/e, the number of electrons per second in a current of 1 A. The current here is 1.6 mA, which gives 10¹⁶.

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