Charge Is Quantized — Question 2

Question 2

Nuclear charge and electron count jointly determine an ion charge

Use $e=1.602176634\times10^{-19}$ C, exact. An ion has 26 protons, 30 neutrons and 23 electrons. Protons carry $+e$, electrons $-e$, and neutrons zero. The ion then loses two electrons without a nuclear change.

1. Find its initial and final net charge in units of $e$ and in coulombs.

2. Compare with an isotope having the same proton and electron counts but 32 neutrons. Determine the electron count required to make either isotope carry charge $-2e$.

Question 2 - Solution

Nuclear charge and electron count jointly determine an ion charge

(a) Method and calculation

$$Q_i=(26-23)e=+3e=4.806529902\times10^{-19}\,\mathrm C.$$$$Q_f=[26-(23-2)]e=+5e=8.010883170\times10^{-19}\,\mathrm C.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{isotope}}=(26-23)e+32(0)=+3e.$$$$26-N_e=-2\Rightarrow N_e=28.$$Changing neutron count changes the isotope and mass, whereas the net electric charge depends on the imbalance between protons and electrons.

Physical check

Electron removal makes the ion more positive; adding neutrons alone does not change its electric charge.