Question

A potential difference is applied between the electrodes in a gas discharge tube. In 1.00 s, 1.90 × 1016 electrons and 1.25 × 1016 singly charged positive ions move in opposite directions through a surface perpendicular to the length of the tube. What is the magnitude of the current in the tube?

Answer 1

Current in the tube will be given by:

I = dQ/dt

dQ = total amount of charge moved = (Np + Ne)*e (Since electron and protons are moving in opposite directions, So total charge will be given by the sum of net charge on both protons and electrons)

dQ = (1.25*10^16 + 1.90*10^16)*1.6*10^-19 = 5.04*10^-3 C

dt = time interval = 1.00 sec

So,

I = 5.04*10^-3 C/1.00 s

I = 5.04*10^-3 Amp = 5.04 mA

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