Question

Use Coulomb's law to estimate the average distance between the sodium nucleus and the 3s electron.

Answer 1

Answer – We are given sodium nucleus and we need to calculate the average distance between the sodium nucleus and the 3s electron.

We know the average distance between the element nucleus is the ionization energy by electrostatic force.

r = E/F , where E- ionization energy and F- electrostatic force.

We also know the Coulomb’s law and it is as follow –

F =k * |q₁|*|q₂| / r²

Where, q₁ = electron charge = -1.6*10^-19­­C , q₂ – sodium nucleus charge

We know, sodium has 11 electrons and 1 e = -1.6*10^19­­C

So, q₂ = 11 * -(-1.6*10^-19­­C)

         = 1.76*10^-18 C

k- Coulomb’s constant = 9.0*10⁹ N.m². C^2-.

Now we know the first ionization energy of sodium and it is 495.8 kJ

We know the ionization energy is in the kJ/mol we need energy in the J/atom, so we need to convert the kJ/mol to J/atm

We know,

1 kJ = 1000 J and 1 mole = 6.023*10²³ atom

So, 495.8 kJ/mol in J/atom as follow –

= 495.8 kJ/mol *1000 J / 1 kJ * 1 mole/ 6.023*10²³ atom

= 8.23*10^-19 J/atom

Now we need to use the Coulomb’s law –

F =k * |q₁|*|q₂| / r²

We know r formula –

r = E/F

Now plugging the F formula in the above one and rearranging the equation for the solving r as follow –

r = E/ ( k * |q₁|*|q₂| / r²)

so, r = k * |q₁|* |q₂| / E

       = 9.0*10⁹ N.m². C^2- * 1.6*10^19­­C * 1.76*10^-18 C / 8.23*10^-19 J.atom^-1

       = 3.078*10^-9 m

So, the average distance between the sodium nucleus and the 3s electron is 3.078*10^-9 m