Question

Complete the following nuclear changes:
$a){}_x^a\text{P}\to Q+{}_0^{-1}\beta$

$b){}_{92}^{328}\text{U}\to {}_{90}^{23}\text{Th}+\cdot \cdot \cdot +energy$

c)$23892P\stackrel{\alpha }{\to }Q\stackrel{\beta }{\to }R\stackrel{\beta }{\to }S$

Answer 1

a) In β decay one neutron decays to one proton with the emission of a beta particle. Hence, in the daughter nucleus the number of protons increases by one and the number of neutrons decreases by one.
$axP\to ax+1P+0-1\beta$

b) Nuclear reaction was accompanied with release of a particle having mass number and atomic number equal to 4 and 2 respectively. Hence emitted atom is an alpha $\left(42\alpha \right)$ particle.
$23892U\to 23490\mathrm{Th}+42\alpha +\mathrm{energy}$

c) Firstly, parent nucleus P changes into daughter nucleus Q due to alpha decay. When an alpha particle is emitted, atomic number and mass number are decreased by 2 and 4 respectively.
$23892P\to 23490Q+42\alpha$

Secondly, parent nucleus Q changes into daughter nucleus R due to beta decay. When a beta particle emits, mass number of the parent nucleus remains the same but atomic number increases by 1.
$23490Q\to 23491R+0-1\beta$
Later, parent nucleus R changes into daughter nucleus S due to beta decay again.
$23491R\to 23492S+0-1\beta$
Complete nuclear reaction is:

$23892P\to 23490Q\to 23491R\to 23492S$