Question

In a Neon discharge tube $2.9\times {10}^{18}\left({Ne}^{+}\right)$ ions move to the right each second, while $1.2\times {10}^{18}$ electrons move to the left per second; electron charge is $1.6\times {10}^{-19}C$. The current in the discharge tube is:

• A. $1A$ toward right
• B. $0.66A$ toward right
• C. $0.66A$ toward left
• D. Zero

Answer 1

The correct option is B $0.66A$ toward right

Given, rate of flow of change, of
$N{e}^{+}$ ions, $I_1=2.9\times {10}^{18}\times 1.6\times {10}^{-19}$ [No. of ions $\times$ change of each ions]

$=0.464A$

of $e^{-}$, $I_2=1.2\times {10}^{18}\times 1.6\times {10}^{-19}$ [conventional current is taken opposite to flow of $e^{-}$]

$=0.192A$

Thus, net current

$I=I_1+I_2$

$=0.464+0.192$

$\Rightarrow I=0.656A$ towards right.