A) \[2\times {{10}^{9}}\]
B) \[2\times {{10}^{19}}\]
C) \[3.2\times {{10}^{19}}\]
D) \[1.6\times {{10}^{18}}\]
Correct Answer: B
Solution :
\[I=\frac{ne}{t}\] \[3.2=\frac{n\times 1.6\times {{10}^{-19}}}{1}\] \[n=\frac{3.2}{1.6\times {{10}^{-19}}}\] \[=2\times {{10}^{19}}\]
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