Question

In a neon discharge tube `2.9 xx 10^(18) Ne^(+)` ions move to the right each second while `1.2 xx 10^(18)` eletrons move to the left per second. Electron charge is `1.6 xx 10^(-9) C`. The current in the discharge tube
A. 0.27 A towards right
B. 0.66 A towards right
C. 0.66 A towards left
D. zero

Answer 1

Correct Answer - B
Given, number of ions, i.e., `n_(1)=2.9xx10^(18)`
`n_(2)=1.2xx10^(18)`
Charge of an electron, i.e., `q=1.6xx10^(-19)C`
So,net electric current, i.e, `l=((n_(1))/(t)+(n_(2))/(t))xxq`
`= ((2.9xx10^(18))/(1)+(1.2xx10^(18))/(1))xx1.6xx10^(-19)=0.66 A`
The net electric current 0.66 towards right.