Question

αdecay producing ²⁰⁸Pb. The parent nuclide is in the decay series produced by ²³²Th, the only naturally occurring isotope of thorium.

Short answer

The α decay equation of ²⁰⁸Pb is – $${}_{84}^{212}P{o}_{128}{\to }_{82}^{208}P{b}_{126}{+}_2^{4}H{e}_2$$.

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

As it is known the equation for α decay is –

$${}_Z^A{X}_{A-Z}{\to }_{z-2}^{A-4}{Y}_{A-Z-2}{+}_2^{4}H{e}_2$$

Where, v_e is the electron's neutrino.

Also, it is known that A = Z + N.

Since, A = 208 and Z = 84.

So, the equation in this case is –

$${}_{84}^{212}P{o}_{128}{\to }_{82}^{208}P{b}_{126}{+}_2^{4}H{e}_2$$

Therefore, the decay equation is $${}_{84}^{212}P{o}_{128}{\to }_{82}^{208}P{b}_{126}{+}_2^{4}H{e}_2$$.