Question

Estimate the binding energy of a KCl molecule by calculating the electrostatic potential energy when the K+ and Cl− ions are at their stable separation of 0.28 nm. Assume each has a charge of magnitude 1.0e.

Answer 1

Answer:

binding energy, E = 0.513 eV

Explanation:

It is given that,

Distance between ions, $r=0.28\ nm=0.28\times 10^{-9}\ m$

Charge, $q=1e=1.6\times 10^{-19}\ C$

Binding energy can be calculated using potential energy as :

$E=\dfrac{kq^2}{r}$

$E=\dfrac{9\times 10^8\times (1.6\times 10^{-19})^2}{0.28\times 10^{-9}}$

$E=8.22\times 10^{-20}\ J$

or

E = 0.513 eV

So, the binding energy of a KCl molecule is 0.513 eV. Hence, this is the required solution.