Question

A copper atom consists of copper nucleus surrounded by 29 electrons. The atomic weight of copper is $63.5 \,g\, mol ^{-1}$. Let us now take two pieces of copper each weighing $10\, g$. Let us transfer one electron from one piece to another for every 100 atoms in that piece. What will be the Coulomb force between the two pieces after the transfer of electrons, if they are $1 \,cm$ apart? Avogadro number $=6 \times 10^{23} mole ^{-1}$, charge on an electron $=-1.6 \times 10^{-19} C$

• A. $1.12 \times 10^{18} N$
• B. $4.24 \times 10^{18} N$
• C. $2.06 \times 10^{18} N$
• D. $5.16 \times 10^{18} N$

Answer 1

Answer: (C)

Here, Avogadro number, $N=6 \times 10^{23} \,mol ^{-1}$ and Atomic weight of copper, $A=63.5 \,g \,mol ^{-1} .$ Therefore, number of atoms in $10 g$ of copper.
$n=\frac{N}{A} \times 10=\frac{6 \times 10^{23} \times 10}{63.5}=9.45 \times 10^{22}$
Since one electron per 100 atoms is transferred from copper piece $A$ to $B$, the charge on copper piece $A$,
$q_{A}=+\frac{n}{100} \times e=+151.2\, C$
Obviously, charge on copper piece $B$,
$q_{B}=-151.2\, C$
Force between the two copper pieces,
$F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} \times q_{B}}{r^{2}}$
Setting $r=1\, cm =0.01 \,m$, we get
$F=2.06 \times 10^{18} N \text { (attractive) }$