In a neon discharge tube $2.9 imes 10^{18}Ne^{+}$ ions move to the right each $second$ while $1.2 imes 10^{18}$ electrons move to the left per $second.$ Electron charge is $1.6 imes 10^{- 19}C.$ The current in the discharge tube

Question

In a neon discharge tube $2.9 	imes 10^{18}Ne^{+}$ ions move to the right each $second$ while $1.2 	imes 10^{18}$ electrons move to the left per $second.$ Electron charge is $1.6 	imes 10^{- 19}C.$ The current in the discharge tube

Dr. Abubacker · Accepted Answer

(b) Net current $i = i_{+} + i_{-} = \frac{(n_{+})(q_{+})}{t} + \frac{(n_{-})(q_{-})}{t}$
$\Rightarrow i = \frac{(n_{+})}{t} 	imes e + \frac{(n_{-})}{t} 	imes e$
$= 2.9 	imes 10^{18} 	imes 1.6 	imes 10^{- 19} + 1.2 	imes 10^{18} 	imes 1.6 	imes 10^{- 19}$
$\Rightarrow i = 0.66\ A$

Q. In a neon discharge tube $2.9 \times 10^{18}Ne^{+}$ 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

Explanation

Reviewed by Dr. Abubacker

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