If \(0.8 \) Ampere current flows through metallic wire for 4 hours, the number of electrons flowi ng through the wire will be

\(7.18 \times 10^{22}\)

The correct answer is option 2. \(7.18 \times 10^{22}\).

The relationship between current (I), charge (Q), and time (t) is given by:

\(Q = I \times t\)

Given,

\(I = 0.8 A\) (current)

\(t = 4 hr\\ t = 4 \text{ hours} \times 3600 \text{ seconds/hour} = 14400 \text{ seconds}\)

Then,

\(\text{Charge, }Q = 0.8 \text{ A} \times 14400 \text{ s} = 11520 \text{ C}\)

Charge of one electron (e):

\(e = 1.602 \times 10^{-19} \text{ C}\)

Thus, the number of electrons,

\(n = \frac{Q}{e}\)

Substituting the values, we get

\(n = \frac{11520 \text{ C}}{1.602 \times 10^{-19} \text{ C/electron}}\\ n = \frac{11520}{1.602 \times 10^{-19}} \approx 7.18 \times 10^{22}\)

The number of electrons flowing through the wire is approximately \(7.18 \times 10^{22}\). Thus, the correct answer is 2. \(7.18 \times 10^{22}\)