In a conducting region, \(10^{19}\) electrons and \(10^{19}\) protons move to the left, while \(10^{19}\) \(\alpha\)-particles move to the right per second. The resulting electric current is \((e = 1.6 \times 10^{-19}\text{ C})\) _________

The electric current is given by the rate of flow of positive charge. The direction of current is the direction of flow of positive charge, and opposite to the direction of flow of negative charge.

Current due to electrons moving left:
\(I_e = n_e e = 10^{19} \times 1.6 \times 10^{-19} = 1.6\) A towards right.

Current due to protons moving left:
\(I_p = n_p e = 10^{19} \times 1.6 \times 10^{-19} = 1.6\) A towards left.

Current due to \(\alpha\)-particles moving right:
\(I_{\alpha} = n_{\alpha} (2e) = 10^{19} \times 2 \times 1.6 \times 10^{-19} = 3.2\) A towards right.

Taking the rightward direction as positive, the net current is:
\(I_{net} = I_e - I_p + I_{\alpha}\)

\(I_{net} = 1.6 - 1.6 + 3.2 = 3.2\) A

The resulting electric current is \(3.2\) A towards right.

Answer: \(3.2\) A towards right