Question

A copper atom has $29$ electrons revolving around the nucleus. A copper ball contains $4 \times {10^{23}}$ atoms. What fraction of the electrons are removed to give the ball a charge of $ + 9.6\mu C$ ?
A.$ \sim 1.8 \times {10^{ - 13}}$
B.$ \sim 1.3 \times {10^{ - 12}}$
C.$ \sim 6 \times {10^{ - 10}}$
D.$ \sim 5.2 \times {10^{12}}$

Answer 1

Hint: Given that the electrons are revolving and are to be removed, it is to be noted that the process of the addition or removal of electrons is known as ionization. The atom or the molecule will acquire a positive or a negative charge on the addition or removal of electrons.

Complete answer:
Step I:
Given that the charge on each copper atom is$n = 29$
Number of electron in each copper atom $e = 4 \times {10^{23}}$
Total charge is given by
$q = n \times e$
Substituting the values, of n and e and evaluating total charge,
$q = 29 \times 4 \times {10^{23}} = 1160 \times {10^{23}}$
Step II:
The ball is to be given a charge of $ + 9.6\mu C$
Or $9.6 \times {10^{ - 6}}C$
Step III:
Charge on one electron$ = 1.6 \times {10^{ - 19}}C$
Total number of electrons in $9.6\mu C = \dfrac{{9.6}}{{1.6 \times {{10}^{ - 19}}}}$
$ = 6 \times {10^{19}}$
Step IV:
Required fraction of electrons that are to be removed to get required charge $ = \dfrac{{6 \times {{10}^{19}}}}{{1160 \times {{10}^{23}}}}$
or \[ = 5.17 \times {10^{12}} \sim 5.2 \times {10^{12}}\]
Step V:
The number of electrons that are to be removed to give the ball a charge of $ + 9.6\mu C \sim 5.2 \times {10^{12}}$

Therefore, Option D is the right answer.

Note:
It is to be noted that in order to add or remove an electron from an atom to give the required charge, a particular minimum amount of energy is to be supplied to remove the electron that is most loosely bounded or is present in the outermost shell of the atom. This energy is called ionization energy. After gaining or losing an electron, the atom becomes an ion. If the size of the atom is greater due to the presence of more shells, the outermost electrons are held less tightly by the nucleus and will be easy to remove. The ionization energy will be less in this case.