Question

Write a balanced nuclear equation for the alpha decay of each of the following radioactive isotopes: a. \({ }_{84}^{208} \mathrm{Po}\) b. \({ }_{90}^{232} \mathrm{Th}\) c. \({ }_{102}^{251} \mathrm{No}\) d. radon-220

Short answer

Alpha decay equations are: \({ }_{84}^{208} \mathrm{Po} \rightarrow { }_{82}^{204} \mathrm{Pb} + { }_{2}^{4} \mathrm{He}\), \({ }_{90}^{232} \mathrm{Th} \rightarrow { }_{88}^{228} \mathrm{Ra} + { }_{2}^{4} \mathrm{He}\), \({ }_{102}^{251} \mathrm{No} \rightarrow { }_{100}^{247} \mathrm{Fm} + { }_{2}^{4} \mathrm{He}\), \({ }_{86}^{220} \mathrm{Rn} \rightarrow { }_{84}^{216} \mathrm{Po} + { }_{2}^{4} \mathrm{He}\).

Step-by-step solution

- Understanding Alpha Decay

In alpha decay, a nucleus loses an alpha particle, which consists of 2 protons and 2 neutrons. This means that the atomic number decreases by 2 and the mass number decreases by 4.

- Decay of \({ }_{84}^{208} \mathrm{Po}\)

The alpha decay of \({ }_{84}^{208} \mathrm{Po}\) results in the emission of an alpha particle (\({ }_{2}^{4}\mathrm{He}\)). The new element formed has an atomic number of 82 and a mass number of 204: \[ \mathrm{{ }_{84}^{208}Po \rightarrow \mathrm{{ }_{82}^{204}Pb} + \mathrm{{ }_{2}^{4}He} } \]

- Decay of \({ }_{90}^{232} \mathrm{Th}\)

The alpha decay of \({ }_{90}^{232} \mathrm{Th}\) involves the emission of an alpha particle. The new nucleus has an atomic number of 88 and a mass number of 228: \[ \mathrm{{ }_{90}^{232}Th \rightarrow \mathrm{{ }_{88}^{228}Ra} + \mathrm{{ }_{2}^{4}He} } \]

- Decay of \({ }_{102}^{251} \mathrm{No}\)

The alpha decay of \({ }_{102}^{251} \mathrm{No}\) releases an alpha particle. The resulting element has an atomic number of 100 and a mass number of 247: \[ \mathrm{{ }_{102}^{251}No \rightarrow \mathrm{{ }_{100}^{247}Fm} + \mathrm{{ }_{2}^{4}He} } \]

- Decay of radon-220

Radon-220, which can be represented as \({ }_{86}^{220} \mathrm{Rn}\), decays by emitting an alpha particle. The new element has an atomic number of 84 and a mass number of 216: \[ \mathrm{{ }_{86}^{220}Rn \rightarrow \mathrm{{ }_{84}^{216}Po} + \mathrm{{ }_{2}^{4}He} } \]

Key concepts

alpha particle emission

Alpha particle emission is a type of nuclear reaction where an unstable nucleus releases an alpha particle to become more stable. An alpha particle consists of 2 protons and 2 neutrons, identical to a helium-4 nucleus.

When an alpha particle is emitted, the original nucleus decreases its atomic number by 2 and its mass number by 4. This change transforms the element into a different one. For example, when ewline polonium-208 ( _{84}^{208}Po) emits an alpha particle (_{2}^{4}He), it becomes lead-204 (_{82}^{204}Pb).

nuclear reactions

Nuclear reactions involve changes in an atom's nucleus and can result in the transformation of one element into another. These reactions emit or absorb large amounts of energy compared to chemical reactions.
Alpha decay, a specific type of nuclear reaction, involves the emission of an alpha particle. This process alters the composition of the nucleus, reducing its atomic number and mass number.

For example, in the alpha decay of thorium-232 (_{90}^{232}Th), an alpha particle (_{2}^{4}He) is emitted, and the nucleus transforms into radium-228 (_{88}^{228}Ra).

radioactive isotopes

Radioactive isotopes, or radioisotopes, are unstable forms of elements that decay over time. These isotopes emit radiation as they transform into more stable forms.
Common types of radiation include alpha particles, beta particles, and gamma rays.

For instance, no Nobelium-251 (_{102}^{251}No) is a radioactive isotope that decays by emitting an alpha particle (_{2}^{}4He), resulting in fermium-247 (_{100}^{247}Fm).

mass number

The mass number of an atom is the total number of protons and neutrons in its nucleus. It is always a whole number and indicates the atom's mass.

During alpha decay, the mass number of the original atom decreases by 4, since the emitted alpha particle carries away 2 protons and 2 neutrons.
For example, radon-220 (_{86}^{220}Rn) decays to form polonium-216 (_{84}^{216}Po) by emitting an alpha particle (_{2}^{4}He), reducing its mass number by 4.

atomic number

The atomic number represents the number of protons in an atom's nucleus and defines the element. Each element has a unique atomic number.

During alpha decay, the atomic number of the original nucleus decreases by 2 because the alpha particle emitted contains 2 protons.
For example, when polonium-208 (_{84}^{208}Po) undergoes alpha decay, it becomes lead-204 (_{82}^{204}Pb), as its atomic number is reduced by 2.