A) \[1.6\times {{10}^{19}}\]
B) \[3.1\times {{10}^{19}}\]
C) \[4.8\times {{10}^{19}}\]
D) \[6.2\times {{10}^{19}}\]
Correct Answer: B
Solution :
The number of\[C{{u}^{2+}}\]ions deposited \[=\frac{q}{2e}=\frac{It}{2e}=\frac{1\times 10}{2\times (1.6\times {{10}^{-19}})}\] \[=\left( \frac{10}{3.2} \right)\times {{10}^{19}}=3.1\times {{10}^{19}}\]
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