Chapter 18, Problem 8P

To determine

The number of electrons leave the battery per minute.

Answer to Problem 8P

The number of electrons $\text{=}\text{2}\text{.109}{\text{×10}}^{\text{21}}\text{electrons/}\text{min}$

Explanation of Solution

Given info:

Voltage, $\text{V=}\text{9}\text{.0}\text{V}$ and resistance, $\text{R= 1}\text{.6}\text{Ω}$

Formula used:

From the Ohm’s law,

The relation between resistance, voltage and current

  $\text{V=IR}$

Where $V=$ voltage, $I=$ current, $R=$ resistance.

After rearranging this expression, $\text{I=}\frac{\text{V}}{\text{R}}$

Calculation:

Substituting all the given values in the formula $\text{I=}\frac{\text{V}}{\text{R}}$

  $\text{=}\frac{\text{9}\text{.0}\text{V}}{\text{1}\text{.6}\text{Ω}}$

  $\text{=}\text{5}\text{.625}\text{A}$

The battery lost $\text{5}\text{.625}$ coulomb of charges in one second.

Charge of an electron is $\text{ 1}\text{.6}{\text{×10}}^{\text{-19}}\text{C}$ , so the charge loss by battery:

Substituting the values and calculate

  $\text{5}\text{.625}\text{A}\text{=}\frac{\text{5}\text{.625}\text{C}}{\text{1}\text{sec}}\text{.}\frac{\text{60}\text{s}}{\text{1}\text{min}}\text{.}\frac{\text{1}\text{electron}}{\text{1}{\text{.6×10}}^{\text{-19}}\text{C}}$

  $\text{=}\text{2}{\text{.109×10}}^{\text{21}}\text{electrons}$

Conclusion:

The number of electrons losses by the battery is $\text{2}{\text{.109×10}}^{\text{21}}\text{electrons}$ .