Question

The mass to charge ratio for ${{\text{A}}^ + }$ ions is $1.97 \times {10^{ - 7}}{\text{kg}}{{\text{C}}^{ - 1}}$. Calculate the mass of atom A.

Answer 1

Hint: An atom that carries an electrical charge is called an ion. If the atom loses electrons, the atom becomes positively charged because the number of positively charged protons will be more than the number of electrons. Electrons are negatively charged and protons are positively charged.

Complete step by step solution:
Atom is the smallest unit of an element. It consists of a central nucleus surrounded by one or more electrons. Nucleus is the central part of an atom. It is composed of protons and neutrons. Nucleus contains most of an atom’s mass. Proton is a positively charged particle. It is found within a nucleus. Ernest was first to introduce protons. Atomic number is equal to the number of protons. The mass of a proton is slightly less than the mass of a neutron. Neutron is an uncharged particle. Protons and neutrons are together known as nucleons. The higher number of neutrons, higher the melting point and higher the boiling point.
It is given that the mass to charge ratio of ${{\text{A}}^ + }$ ions, ${\text{r}}$$ = 1.97 \times {10^{ - 7}}{\text{kg}}{{\text{C}}^{ - 1}}$
The ${{\text{A}}^ + }$ ion has a single positive charge. It is equal to the charge of a proton.
Charge of ${{\text{A}}^ + }$ ion, ${\text{c}} = 1.6 \times {10^{ - 19}}{\text{C}}$
Thus, mass of A atom, ${\text{m = r}} \times {\text{c}} \Leftrightarrow {\text{m = 1}}{\text{.97}} \times {\text{1}}{{\text{0}}^{ - 7}}{\text{kg}}.{{\text{C}}^{ - 1}} \times 1.6 \times {10^{ - 19}}{\text{C = 3}}{\text{.52}} \times {\text{1}}{{\text{0}}^{ - 26}}{\text{kg}}$

So the mass of atom A is ${\text{3}}{\text{.52}} \times {\text{1}}{{\text{0}}^{ - 26}}{\text{kg}}$

Note: Mass to charge ratio is different from charge to mass ratio of electron which is calculated from Thomson’s cathode ray discharge tube experiment. The charge to mass ratio of electron is found to be $1.76 \times {10^{11}}{\text{C}}.{\text{k}}{{\text{g}}^{ - 1}}$. He applied electrical and magnetic fields perpendicular to each other as well as the path of electrons.