# Find, to three significant digits, the charge and the mass of the following particles: a....

## Question:

Find, to three significant digits, the charge and the mass of the following particles:

a. quadruply ionized nitrogen atoms ({eq}N^{4+} {/eq}) found in plasma in a hot star,

b. the nucleus of a nitrogen atom and

c. the molecular ion, {eq}H_2O^{-} {/eq}.

## Ions

Ions are the individual atoms or molecule that has an excess of electrons or deficiency of electrons, The ions having an excess of electrons are known as negatively charged ions or anions and the ions having the deficiency of electrons are positively charged ions or cations

Part A)

Charge on the ionized nitrogen atom as it has n= 4 excess protons or deficiency of 4 electrons

{eq}\begin{align} q_{N^{4+}} = ne \\ q_{N^{4+}} = 4 \times 1.6 \times 10^{-19} \ \rm C \\ q_{N^{4+}} = 6.40 \times 10^{-19} \ \rm C \end{align} {/eq}

Mass of the ionized nitrogen will be same as the mass of the nitrogen atom as the mass of the electron's is almost neglible.

{eq}\begin{align} m_{N^{4+}} = 14 \ \rm u \\ m_{N^{4+}} = 14 \times 1.66 \times 10^{-27} \ \rm kg \\ m_{N^{4+}} = 2.32 \times 10^{-26} \ \rm kg \end{align} {/eq}

Part B) Charge on the nitrogen nucleus as it has n= 7 excess protons or deficiency of 7 electrons

{eq}\begin{align} q_{N^{7+}} = ne \\ q_{N^{7+}} = 7 \times 1.6 \times 10^{-19} \ \rm C \\ q_{N^{7+}} = 1.12 \times 10^{-18} \ \rm C \end{align} {/eq}

Mass of the ionized nitrogen will be same as the mass of the nitrogen atom as the mass of the electron's is almost neglible.

{eq}\begin{align} m_{N^{7+}} = 14 \ \rm u \\ m_{N^{7+}} = 14 \times 1.66 \times 10^{-27} \ \rm kg \\ m_{N^{7+}} = 2.32 \times 10^{-26} \ \rm kg \end{align} {/eq}

Part C) Charge on the ionized water molecule as it has n= 1 excess electrons

{eq}\begin{align} q_{H_2O^{-1}} = ne \\ q_{H_2O^{-1}} = 1 \times 1.6 \times 10^{-19} \ \rm C \\ q_{H_2O^{-1}} = 1.60 \times 10^{-19} \ \rm C \end{align} {/eq}

Mass of the ionized water molecule

{eq}\begin{align} m_{H_2O^{-1}} = 2 m_{H} + m_{O} \\ m_{H_2O^{-1}} = 2 \times 1 \ \rm u + 16 \ \rm u \\ m_{H_2O^{-1}} = 18 \ \rm u \\ m_{H_2O^{-1}} = 18 \times 1.66 \times 10^{-27} \ \rm kg \\ m_{H_2O^{-1}} = 2.99 \times 10^{-26} \ \rm kg \end{align} {/eq} 