Repulsion energy of two atoms

[w3390]

Homework Statement

The dissociation energy of a hypothetical molecule AB (composed of A+ and B- ions) is 5.24 eV, and the equilibrium separation of AB is 0.216 nm. The electron affinity of a B atom is 3.36 eV and the ionization energy of element A is 4.00 eV. Determine the core-repulsion energy of AB.

Homework Equations

U(rep)=-[E(diss)+deltaE+Ue], where delta E is the dissociation energy minus the electron affinity

Ue=-(ke^2)/r

The Attempt at a Solution

First I found delta E. Delta E= 4.00ev-3.36ev=.64eV.

The dissociation energy, E(diss), is given as 5.24eV.

I calculated the electrostatic potential energy and converted it to electron volts and got -6.6472eV.

Therefore, U(rep)=-[5.24eV+.64eV-6.6472eV]=.767eV

However, my assignment is telling me this is incorrect. Can anyone find where I went wrong? Any help is greatly appreciated.

 

[Jhero]

Hi there, based on your attempt at solving this problem, it looks like you have the right idea but there are a few minor mistakes in your calculations. Here is a step-by-step solution to help you find where you went wrong:

1. First, you correctly calculated delta E as 0.64 eV. This is the difference between the dissociation energy (5.24 eV) and the electron affinity of B (3.36 eV).

2. Next, you need to calculate the electrostatic potential energy, Ue, using the formula Ue = -(ke^2)/r. Be careful to use the correct units for this calculation. The value for the Coulomb constant, k, is 8.99 x 10^9 Nm^2/C^2. The equilibrium separation, r, is given in nanometers (nm), so you will need to convert it to meters (m) before plugging it into the formula.

Ue = -(8.99 x 10^9 Nm^2/C^2)(1.602 x 10^-19 C)^2/(0.216 x 10^-9 m) = -6.6472 x 10^-19 J

To convert this to electron volts (eV), divide by the elementary charge (1.602 x 10^-19 C).

Ue = (-6.6472 x 10^-19 J)/(1.602 x 10^-19 C) = -6.6472 eV

3. Now, you can plug in the values for E(diss), delta E, and Ue into the formula for U(rep).

U(rep) = -[5.24 eV + 0.64 eV - (-6.6472 eV)] = -0.2332 eV

Therefore, the correct answer for the core-repulsion energy of AB is -0.2332 eV.

I hope this helps! Remember to always double check your units and be careful with your calculations. Good luck with your future studies!