question_answer

In an electrolyte \[3.2\times {{10}^{18}}\] bivalent positive ions drift to the right per second while \[3.6\times {{10}^{18}}\] monovalent negative ions drift to the left per second. Then the current is

A)                    1.6 amp to the left     

B)            1.6 amp to the right

C)                    0.45 amp to the right

D)                    0.45 amp to the left

Correct Answer: B

Solution :

             Net current \[{{i}_{net}}={{i}_{(+)}}+{{i}_{(-)}}\] \[=\frac{{{n}_{(+)}}{{q}_{(+)}}}{t}+\frac{{{n}_{(-)}}{{q}_{(-)}}}{t}\] \[=\frac{{{n}_{(+)}}}{t}\times 2e+\frac{{{n}_{(-)}}}{t}\times e\] \[=\text{ }3.2\times {{10}^{18}}\times 2\times 1.6\times {{10}^{19}}+\text{ }3.6\times {{10}^{18}}\times 1.6\times {{10}^{19}}\] = 1.6 A (towards right)