In a neon discharge tube 2.9 × 1018 Ne ions move to the right each second, while 1.2 × 1018 electrons move to the left per sec, electron charge is 1.6 × 10-19 C. The current in the discharge tube is

Question

In a neon discharge tube 2.9 × 1018 Ne ions move to the right each second, while 1.2 × 1018 electrons move to the left per sec, electron charge is 1.6 × 10-19 C. The current in the discharge tube is (A) 1 A towards right (B) 0.66 A towards right (C) 0.66 A towards left (D) zero

Tardigrade Admin · Accepted Answer

In the two cases current is flowing in the right direction, therefore total current I=IN e++Ie or I=(NN e++Ne) e or I=(2.9 × 1018+1.2 × 1018) × 1.6 × 10-19 ⇒ I=0.66 A A towards right

Q. In a neon discharge tube $2.9 \times 1 0^{18} N e$ ions move to the right each second, while $1.2 \times 1 0^{18}$ electrons move to the left per sec, electron charge is $1.6 \times 1 0^{- 19} C$ . The current in the discharge tube is

1 A towards right

0.66 A towards right

0.66 A towards left

zero

In the two cases current is flowing in the right direction,
therefore total current $I = I_{N e^{+}} + I_{e}$
or $I = (N_{N e^{+}} + N_{e}) e$ or
$I = (2.9 \times 1 0^{18} + 1.2 \times 1 0^{18}) \times 1.6 \times 1 0^{- 19}$
$\Rightarrow I = 0.66 A$ A towards right

Q. In a neon discharge tube 2.9×1018Ne ions move to the right each second, while 1.2×1018 electrons move to the left per sec, electron charge is 1.6×10−19C. The current in the discharge tube is