Question

Identify each of the following:

a. ${}_{-1}{}^0\mathrm{X}$

b. ${}_2{}^4\mathrm{X}$

c. ${}_0{}^1\mathrm{X}$

d. ${}_{18}{}^{38}\mathrm{X}$

e.${}_6{}^{14}\mathrm{X}$

Short answer

a. It is a beta particle $\left(\mathrm{\beta }\right)$

b. It is an alpha particle $\left(\mathrm{\alpha }\right)$

c. It is a neutron $\left({}_0{}^1\mathrm{n}\right)$

d. It is ${}_{18}{}^{38}\mathrm{Ar}$

e. It is${}_6{}^{14}\mathrm{C}$

Step-by-step solution

Part (a) Given Information

We need to identify this particle:${}_{-1}{}^0\mathrm{X}$

Part (b) Explanation

A beta particle is a high-energy electron and it has a charge of $1-$, and because its mass is very less than the mass of a proton, it has a mass number of zero. It is represented by the $\left(\mathrm{\beta }\right)$.

Thus, ${}_{-1}{}^0\mathrm{X}$is a beta particle. It is represented as${}_{-1}{}^0\mathrm{e}$.

Part (b) Given Information

We need to identify this particle :${}_2{}^4\mathrm{X}$

Part (b) Explanation

Analpha particleis similar to a helium (He) atom, which has 2 protons and 2 neutrons. An alpha particle has an atomic mass of four, an atomic number of two, and a charge of. It is represented as alpha $\left(\mathrm{\alpha }\right)$or the symbol of a helium nucleus except that the 2+ charge is omitted.

Thus, ${}_2{}^4\mathrm{X}$is an alpha particle, where X is$\mathrm{He}$. It can be represented as${}_2{}^4\mathrm{He}$

Part (c) Given Information

We need to identify this particle:${}_0{}^1\mathrm{X}$

Part (c) Explanation

This particle is known as a neutron. It is represented as ${}_0{}^1\mathrm{n}$. The mass number for neutron is 1 and it has a zero charge.

Thus, ${}_0{}^1\mathrm{X}$is a neutron and it can be shown as${}_0{}^1\mathrm{n}$.

Part (d) Given Information

We need to identify this particle:${}_{18}{}^{38}\mathrm{X}$

Part (d) Explanation

The given particle has the atomic number 18 while the mass number is 38. The element having atomic number 18 is Argon which means it has 18 protons and 18 electrons.

Thus, ${}_{18}{}^{38}\mathrm{X}$can be written as ${}_{18}{}^{38}\mathrm{Ar}$.

Part (d) Given Information

We need to identify this particle:${}_6{}^{14}\mathrm{X}$

Part (d) Explanation

The given particle has the atomic number 6 and the mass number for this particle is 14. The element with atomic number 6 is Carbon which means it has 6 protons and 6 electrons.

Thus,${}_6{}^{14}\mathrm{X}$can be written as${}_6{}^{14}\mathrm{C}$.