Answer
$1.00\times10^{19}\frac{e^-}{s}$
Work Step by Step
Use the definition of current, Eq. 18–1.
$$I=\frac{\Delta Q}{\Delta t}$$
$$1.60A=\frac{1.60C}{s}\frac{1\;e^-}{1.60\times10^{-19}C}=1.00\times10^{19}\frac{e^-}{s}$$
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