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Question: In a neon discharge tube \(2.9 \times {\text{1}}{{\text{0}}^{18}}(N{e^ + })\) ions move to the right...

In a neon discharge tube 2.9×1018(Ne+)2.9 \times {\text{1}}{{\text{0}}^{18}}(N{e^ + }) ions move to the right each second, while 1.2×10181.2 \times {10^{18}} electrons move to the left per second; electron charge is 1.6×10−19C1.6 \times {10^{ - 19}}C. The current in the discharge tube is:
A. 1A toward right B. 0.66A toward right C. 0.66A toward left D. zero {\text{A}}{\text{. 1A toward right}} \\\ {\text{B}}{\text{. 0}}{\text{.66A toward right}} \\\ {\text{C}}{\text{. 0}}{\text{.66A toward left}} \\\ {\text{D}}{\text{. zero}} \\\

Explanation

Solution

Hint: Find the net current flowing in the tube, caused by the positive ions and the electrons. To get the direction of current, remember that current flows in the opposite direction of the motion of electrons.

Complete step-by-step solution -
Formula used: I = (nq / t) ; where n is the number of electrons or ions, q is the charge on an electron and t is time.
Total charge = I1+I2I_1 + I_2.
Current is the rate of charge per unit time.
Current flowing because of positive ions (I1)(I_1):
=2.9×1018×1.6×10−19 =.464A  = 2.9 \times {10^{18}} \times 1.6 \times {10^{ - 19}} \\\ = .464A \\\
Towards right (as current flows in direction of positive charges)
Current flowing because of electrons (I2)(I_2):
=1.2×1018×1.6×10−19 =.192A  = 1.2 \times {10^{18}} \times 1.6 \times {10^{ - 19}} \\\ = .192A \\\
Towards right (as current flows in opposite direction of flow of electron)
Total current = (.464+.192)A = (0.666) Amperes
So, the correct option is B.

Note: Ne+ has +1 charge so in the formula q=1×1.6×10−191 \times 1.6 \times {10^{ - 19}}has been put if it had been Ne +2 q=2×1.6×10−192 \times 1.6 \times {10^{ - 19}}would have been put. It is important to put the direction of the current correctly. Since current is a scalar quantity, simple addition and subtraction is done.