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Inside a chemical battery it is not actually individual electrons that are transported from the + end to the 鈥� end. At the + end of the battery an 鈥渁cceptor鈥� molecule picks up an electron entering the battery, and at the 鈥� end a different 鈥渄onor鈥� molecule gives up an electron, which leaves the battery. Ions rather than electrons move between the two ends to support the charge inside the battery.

When the supplies of acceptor and donor molecules are used up in a chemical battery, the battery is dead because it can no longer accept or electron. The electron current in electron per second times the number of seconds of battery life, is equal to the number of donor molecules in the battery.

A flashlight battery contains approximately half a mole of donor molecules. The electron current through a thick filament bulb powered by two flashlight batteries in series is about 0.3 A. About how many hours will the batteries keep this bulb lit?

Step by step solution

01

Identification of given data

The electron current through batteries is$I=0.3\text{A}$

The number of donor molecule is $n=0.5N_A$

The number of hours is found by equating the total charge due to donor molecules divided by the electron current of batteries.

02

Determination of expression for number of hours batteries keep bulb lit

The number of hours of batteries keeps lit is given as:$$\begin{gathered} t=\frac{ne}{I} \\ t=\frac{0.5N_{A}e}{I} \end{gathered}$$

Here, $N_A$is the Avogadro鈥檚 constant and its value is $6.023脳{10}^{23}$, is the charge of an electron and its value is$1.6脳{10}^{-19}\text{C}$ .

03

Determination of number of hours batteries keep bulb lit

Substitute all the values in above equation.

$$\begin{gathered} t=\frac{0.5\left(6.023脳{10}^{23}\right)\left(1.6脳{10}^{-19}\text{C}\right)}{0.3\text{A}} \\ t=160613.3\text{s} \\ t=\left(160613.3\text{s}\right)\left(\frac{1\text{h}}{3600\text{s}}\right) \\ t鈮�45\text{h} \end{gathered}$$

Therefore, the number of hours of batteries keeps lit is .

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