Question

How much energy will be released when a sodium ion and a chloride ion, originally at infinite distance are brought together to a distance of 2.76 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] (the shortest distance of approach in a sodium chloride crystal)? Assume that ions act as point charges, each with a magnitude of \[\text{1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-19}}}\text{C}\]
Permittivity constant of the medium is \[\text{9 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{9}}}\text{N}{{\text{m}}^{\text{2}}}{{\text{C}}^{\text{-2}}}\]

Answer 1

Hint: Since the chlorine and sodium ion are considered to be point charges, we can apply the Coulomb’s law. This law can be applied for any two point charges which have a fixed distance between atoms, as here we have a sodium ion and chlorine ion.

Complete answer:
Let us try to formulate the numerical given here
- Now, when two point charges are brought in close proximity of each other, they combine together and release an enormous amount of energy.
-Since sodium is positively charged and chlorine molecules are negatively charged, both when combined to form NaCl, they emit energy, which is denoted by a negative sign.
-As we know that the two point charges will obey the electrostatic law of attraction, we can use the formula as:
Force between two point charges ${{q}_{1}}$ and ${{q}_{2}}$ is given by: \[{\text{k}{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}/{{{\text{r}}^{\text{2}}}}\;\]
So, the energy between the charges is given by: \[{\text{k}{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}/{\text{r}}\;\]
Where k is the permittivity constant and
  ${{q}_{1}}$ and ${{q}_{2}}$ are the charges on the sodium and chloride ion respectively
And r is the distance between the ions.
But as energy is released,so we are applying a negative sign: :- \[{\text{k}{{\text{q}}_{\text{1}}}{{\text{q}}_{\text{2}}}}/{\text{r}}\;\]
So ,now applying the formula in the question given above
Since both ions carry the same charge \[{{\text{q}}_{\text{1}}}\text{=}{{\text{q}}_{\text{2}}}\text{=1}\text{.6 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-19}}}\text{C}\]
Distance is between the ions: 2.76 \[\overset{\text{o}}{\mathop{\text{A}}}\,\]
= \[\text{2}\text{.76 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-10}}}\text{m}\]\[\]
Therefore we get = -\[\text{8}\text{.36 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-19}}}\text{J}\] ( required energy)

Note:
Remember we are trying to find the energy released, so a negative sign is a must. So,in order to avoid errors during calculating,we should keep in mind to convert the quantities in standard units.