Solved Problem on Electric Current
advertisement   



Suppose that in an experiment of electrochemistry, you remove 1 electron from each set of 10 atoms in a block of copper of mass m = 0.3 kg. The molar mass of copper is 64 g/mol. Determine the total charge removed from the block.
Data: Avogadro's number \( N_{\small A}=6.02\times 10^{23} \), the elementary charge of an electron \( -1.6\times 10^{-19}\;\mathrm C \).


Problem data:
  • Mass of block of copper:    m = 0.3 kg;
  • Molar mass of copper:    M = 64 g/mol;
  • Fraction of electrons removed:    \( \displaystyle f=\frac{1}{10}\;\frac{\text{electrons}}{\text{atom}} \);
  • Elementary charge of the electron:    \( e=-1.6\times 10^{-19}\;\mathrm C \);
  • Avogadro's Number:    \( N_{\small A}=6.02\times 10^{23} \).
Solution

First, we must convert the mass of copper given in kilograms (kg) to grams (g)
\[ \begin{gather} m=0.3\;\mathrm{\cancel{kg}}\times\frac{1000\;\mathrm g}{1\;\mathrm{\cancel{kg}}}=300\;\mathrm g \end{gather} \]
The total electric charge of a body is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=n_e e} \tag{I} \end{gather} \]
where ne is the number of electrons in the sample.
The number of moles, n, of a given mass is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {n=\frac{m}{M}} \end{gather} \]
\[ \begin{gather} n=\frac{300\;\mathrm{\cancel g}}{64\;\mathrm{\frac{\cancel g}{mol}}}\\[5pt] n\approx 4.7\;\text{moles of atoms of copper} \end{gather} \]
The number of copper atoms, na, contained in the block will be
\[ \begin{gather} n_a=nN_A\\[5pt] n_a=\left(4.7\;\mathrm{\cancel{moles}}\text{ of atoms of copper}\right)\left(6.02\times 10^{23}\;\frac{1}{\mathrm{\cancel{mol}}}\right)\\[5pt] n_a\approx2.8\times 10^{24}\;\text{atoms of copper} \end{gather} \]
As we remove 1 electron from every 10 atoms, the total number of electrons removed from the block will be
\[ \begin{gather} n_e=n_a f\\[5pt] n_e=\left(2.8\times 10^{24}\;\cancel{\text{atoms}}\right)\left(\frac{1}{10}\;\frac{\text{electrons}}{\cancel{\text{atoms}}}\right)\\[5pt] n_e=2.8\times 10^{23}\;\text{electrons} \end{gather} \]
substituting this value in expression (I)
\[ \begin{gather} Q=\left(2.8\times 10^{23}\;\cancel{\text{electrons}}\right)\left(-1.6\times 10^{-19}\;\frac{\text{C}}{\cancel{\text{electrons}}}\right) \end{gather} \]
\[ \begin{gather} \bbox[#FFCCCC,10px] {Q\approx -4.5\times 10^4\;\text{C}} \end{gather} \]
advertisement   

Creative Commons License
Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .