Question

A certain nucleus A (mass number 238 and atomic number 92) is radioactive and becomes a nucleus B (mass number 234 and atomic number 90) by the emission of a particle.
a) Name the particle emitted.
b) Explain how you arrived at your answer.
c) State the change in the form of a reaction.

Answer 1

(a) The particle emitted is the Alpha $\left(\alpha \right)$ particle.
(b) Let the emitted particle is X with atomic number Z and mass number A, such that
${}_{92}^{238}\text{A}{\to }_{90}^{234}\text{B}{+}_Z^A\text{X}$
Equating mass number in above equation, Mass numbers of nuclei A =
Sum of mass numbers of nuclei B and X
238 = 234 + A
$\Rightarrow$A = 4
Similarly, on equating the atomic number in above equation,
92 = 90 + Z
⇒ Z = 2
Here atomic number (Z) and mass number (A) of emitted particles are 2 and 4. It represents an alpha particle emission.
Here X is ${}^{{}_2^4\text{He}}$.
(c) Radioactive decay of nucleus A can be represented as
${}_{92}^{238}\text{A}{\to }_{90}^{234}\text{B}{+}_2^4\text{He}$