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Chapter 17: Q17.2E (page 957)

For the scenario in the previous question, how many electrons moved through the circuit.

Short Answer

Expert verified

Number of electrons is \({\rm{3}}{\rm{.28 \times 1}}{{\rm{0}}^{{\rm{22}}}}\)

Step by step solution

01

Formula of number of electron

The charge of 1 electron is equal to\({\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}{\bf{C}}\). Thus, in 1C, the number of electrons will be:

\(\begin{array}{l}{\bf{1C = }}\frac{{\bf{1}}}{{{\bf{1}}{\bf{.6022 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}}}{\bf{ }}\\{\bf{electrons = 6}}{\bf{.2414 \times 1}}{{\bf{0}}^{{\bf{18}}}}{\bf{ electrons}}\end{array}\)

02

Determine the number of electrons

For 5250 C of charge, the number of electrons can be calculated as follows:

\({\rm{1C}} \to {\rm{6}}{\rm{.2414 \times 1}}{{\rm{0}}^{{\rm{18}}}}{\rm{ electrons }}\)

Thus, \(5250{\rm{C}} \to (5250)\left( {6.2414 \times {{10}^{18}}{\rm{ electrons }}} \right) = 3.28 \times {10^{22}}{\rm{ electrons }}\)

Therefore, the number of electrons are \(3.28 \times 1{0^{22}}\).

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