Question

The batteries of a submerged non-nuclear submarine supply 1000 A at full speed ahead. How long does it take to move Avogadro’s number 6.023 x 10²³ of electrons at this rate?

Short answer

It takes \(96.32\;{\rm{s}}\)to move Avogadro’s number of electrons at this rate.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The current supplied by the submarine is\(I = 1000\;{\rm{A}}\).
• The value of the Avogadro’s number is \(N = 6.02 \times {10^{23}}\;{\rm{electron}}\).

Significance of the Avogadro’s number

The Avogadro’s number is described as the number of molecules in one substance. Moreover, it is also referred to as the number of units in one mole.

Determination of the time

The equation of the time taken to move Avogadro’s number of electrons is expressed as:

\(t = \frac{{N{Q_e}}}{I}\)

Here,\(t\)is the time taken to move Avogadro’s number of electrons,\(N\)is the value of the Avogadro’s number,\({Q_e}\)is the charge of one electron, and\(I\)is the current supplied by the submarine.

Substitute the values in the above equation.

\(\begin{array}{c}t &= \frac{{\left( {6.02 \times {{10}^{23}}\;{\rm{electron}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}/{\rm{electron}}} \right)}}{{1000\;{\rm{A}}}}\\ & = \frac{{\left( {96320\;{\rm{C}}} \right)}}{{1000\;{\rm{A}}}}\\ & = 96.32\;{\rm{C}} \cdot {\rm{A}} \times \frac{{1\;{\rm{A}} \cdot {\rm{s}}}}{{1\;{\rm{C}}}}\\ & = 96.32\;{\rm{s}}\end{array}\)

Thus, it takes \(96.32\;{\rm{s}}\) to move Avogadro’s number of electrons at this rate.