Solved Problem on Electric Current
advertisement   



Suppose that from each group of s atoms of a metal sample, it is possible to remove f electrons. Consider a block of mass m. Assume M is the molecular mass of the metal. What will be the charge of this block if we remove f electrons?


Problem data:
  • Mass of block:    m;
  • Molecular mass of metal:    M;
  • Electron fraction removed:    \( \displaystyle \frac{f}{s}\;\frac{\text{electrons}}{\text{atoms}} \).
Solution

The total electric charge of a body is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {Q=n_{e}\mathrm e} \tag{I} \end{gather} \]
where ne is the number of electrons in the sample and e is the elementary charge.
The number of moles n of a mass is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {n=\frac{m}{M}} \tag{II} \end{gather} \]
The number of atoms na in the copper contained in the block will be
\[ \begin{gather} n_{a}=nN_{A} \tag{III} \end{gather} \]
where NA is the number of Avogadro, substituting the expression (III) in (II), we have
\[ \begin{gather} n_{a}=\frac{m}{M}N_{A}\;\text{atoms} \tag{IV} \end{gather} \]
As we removed f electrons of each s atoms, the total number of electrons removed from the block will be
\[ \begin{gather} n_{e}=n_{a}\frac{f}{s} \tag{V} \end{gather} \]
substituting the expression (IV) in (V), we have
\[ \begin{gather} n_{e}=\left(\frac{m}{M}N_{A}\;\cancel{\text{atoms}}\right)\times \left(\frac{f}{s}\;\frac{\text{electrons}}{\cancel{\text{atoms}}}\right)\\ n_{e}=\frac{mN_{A}}{M}\frac{f}{s}\;\text{electrons} \tag{VI} \end{gather} \]
substituting the expression (VI) in (I), we obtain
\[ Q=\frac{mN_{A}}{M}\frac{f}{s}e \]
\[ \bbox[#FFCCCC,10px] {Q=\frac{mefN_{A}}{sM}} \]
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 .