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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 tubeA. 0.27 A towards rightB. 0.66 A towards rightC. 0.66 A towards leftD. zero |
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Answer» 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. |
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