## Physics (10th Edition)

(a) $1.5\times10^{-11}\,A$ (b) $4.7\times10^{7}\,ions$
(a) According to Ohm's law, $I=\frac{V}{R}=\frac{75\times10^{-3}\,V}{5.0\times10^{9}\,\Omega}=1.5\times10^{-11}\,A$ (b) $I=\frac{q}{t}\implies q=It=1.5\times10^{-11}\,A\times0.50\,s$ $=7.5\times10^{-12}\,C$ Let the number of $Na^{+}$ ions be n. Then, $n=\frac{q}{+e}=\frac{7.5\times10^{-12}\,C}{1.6\times10^{-19}\,C/ion}=4.7\times10^{7}\,ions$