Question

α decay producing ²²⁸Ra. The parent nuclide is nearly 100% of the natural element and is found in gas lantern mantles and in metal alloys used in jets ²²⁸Ra is also radioactive.

Short answer

The α decay equation of \[{}_{88}^{228}Ra\] is – \[_{90}^{232}T{h_{142}} \to _{88}^{228}R{a_{140}} + _2^4H{e_2}\].

Step-by-step solution

Concept Introduction

The process of nuclear decay in which the parent nucleus emits an alpha particle is known as alpha decay. Two protons and two neutrons make up the alpha particle, which is structurally comparable to the nucleus of a helium atom and is symbolised by the Greek letter α.

Finding α decay equation

α decay means emission of α particleand α particle is the nucleus of helium atom containing 2 neutrons and 2 protons. So, general equation can be written as –

\[_Z^A{X_{A - Z}} \to _{z - 2}^{A - 4}{Y_{A - Z - 2}} + _2^4H{e_2}\]

Since \[{}_{88}^{228}Ra\] has atomic number 88.So, before alpha decay, parent nucleus will have atomic number 88 + 2 = 90. Now, the atomic mass of \[{}_{88}^{228}Ra\] is 228 so before alpha decay the parent nucleuswill have atomic mass 228 + 4 = 232. Hence,the parent nucleus is \[{}_{90}^{232}Th\].So,the decay equation will be –

\[_{90}^{232}T{h_{142}} \to _{88}^{228}R{a_{140}} + _2^4H{e_2}\]

Therefore, the decay equation is \[_{90}^{232}T{h_{142}} \to _{88}^{228}R{a_{140}} + _2^4H{e_2}\].