Question

Complete the following nuclear changes:
a) ${}_x^a\text{P}\to Q{+}_{-1}^0\beta$
b) ${}_{92}^{238}\text{U}{\to }_{90}^{234}\text{Th}+\dots +energy$
c) ${}_{92}^{238}\text{P}{\to }^{\alpha }Q{\to }^{\beta }R{\to }^{\beta }S$

Answer 1

a) In $\beta$ 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 by one and the number of neutron decreases by one.
${}_x^a\text{P}{\to }_{x+1}^a\text{P}{+}_{-1}^0\beta$

b) Nuclear rection 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 ${(}_2^4\alpha )$ particle.
${}_{92}^{238}\text{U}{\to }_{90}^{234}\text{Th}{+}_2^4\alpha \dots +energy$

c) Firstly parent nucleus $P$ changes into daughter nucleus Q due to alpha decay. When an alpha particle is emmited, atomic number and mass number are decreased by 2 and 4 respectively.
${}_x^a\text{P}{\to }_{x+1}^a\text{P}{+}_{-1}^0\beta$
Secondly parent nucleus $Q$ changes into daughter nucleus $R$ due to beta decay. When a beta particle is emmited, atomic number increses by 1 but mass number of the parent nuclei remains same.
${}_{90}^{234}\text{Q}{\to }_{91}^{234}\text{R}{+}_{-1}^0\beta$
Later parent nucleus changes into daughter nuclei $S$ due to bet decay agin.
${}_{91}^{234}\text{R}{\to }_{92}^{234}\text{S}{+}_{-1}^0\beta$

Therefore the complete nuclear reaction is
${}_{92}^{238}\text{P}{\to }^{\alpha }Q{\to }^{\beta }R{\to }^{\beta }S$