Question Number 10250 by jaikar last updated on 31/Jan/17
Commented by jaikar last updated on 31/Jan/17
$${please}\:{Q}\:{no}.\:\mathrm{10} \\ $$$$ \\ $$
Answered by ridwan balatif last updated on 31/Jan/17
$$\mathrm{10}.\:\mathrm{n}=\mathrm{1}.\mathrm{5625}×\mathrm{10}^{\mathrm{18}} \mathrm{electron},\:\mathrm{1}\:\mathrm{electron}\:\mathrm{has}\:\approx\:\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} \mathrm{C} \\ $$$$\mathrm{Q}/\mathrm{t}=\mathrm{n}.\mathrm{e}/\mathrm{1s} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}.\mathrm{5625}×\mathrm{10}^{\mathrm{18}} ×\mathrm{1}.\mathrm{6}×\mathrm{10}^{−\mathrm{19}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{1}} \mathrm{C}/\mathrm{s} \\ $$$$\mathrm{t}\:\:=\frac{\mathrm{12C}}{\mathrm{2}.\mathrm{5}×\mathrm{10}^{−\mathrm{1}} \mathrm{C}/\mathrm{s}}=\mathrm{48s}\: \\ $$$$\mathrm{Answer}:\:\mathrm{C} \\ $$
Commented by jaikar last updated on 31/Jan/17
$${thanks} \\ $$$$ \\ $$