Some Properties of Atoms — Question 5

Question 5

Successive ionization energies add along a charge-state path

A model atom has successive ionization energies $I_1=5.0\,\mathrm{eV}$, $I_2=12.0\,\mathrm{eV}$ and $I_3=28.0\,\mathrm{eV}$. Initially the neutral atom is in its ground state; removed electrons have negligible final kinetic energy.

1. Find the minimum work to create the doubly and triply charged ions.

2. A 50.0 eV energy deposit creates the triply charged ground-state ion with negligible recoil. Find the total outgoing electron kinetic energy.

Question 5 - Solution

Successive ionization energies add along a charge-state path

(a) Method and calculation

$$W_{0\to2+}=I_1+I_2=\boxed{17.0\,\mathrm{eV}}.$$$$W_{0\to3+}=I_1+I_2+I_3=\boxed{45.0\,\mathrm{eV}}.$$

(b) Interpretation and further calculation

$$E_{\mathrm{deposit}}=W_{0\to3+}+\sum_{j=1}^3K_j.$$$$\boxed{\sum K_j=50.0-45.0=5.0\,\mathrm{eV}}.$$

Physical check

Successive ionization energies refer to different initial charge states. The third ionization energy alone is not the work to remove three electrons from a neutral atom.