In the nuclear decay given below: 
${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}\to {}_{\mathrm{Z}+1}{}^{\mathrm{A}}\mathrm{Y}\to {}_{\mathrm{Z}-1}{}^{\mathrm{A}-4}\mathrm{B}\to {}_{\mathrm{Z}-1}{}^{\mathrm{A}-4}\mathrm{B}$
the particles emitted in the sequence are:

A nuclear reaction is given by

${}_{\mathrm{Z}}{\mathrm{X}}^{\mathrm{A}}{\to }_{\mathrm{Z}+1}{\mathrm{Y}}^{\mathrm{A}}{+}_{-1}{\mathrm{e}}^0+\overline{\mathrm{v}}$

represents:

1. β-decay

2. γ-decay

3. fusion

4. fission

An atomic nucleus ${}_{90}{}^{232}\mathrm{Th}$ emits several $\mathrm{\alpha }$ and $\mathrm{\beta }$ radiations and finally reduces to ${}_{82}{}^{208}\mathrm{Pb}$. It must have emitted:
1. 4$\mathrm{\alpha }$ and 2$\mathrm{\beta }$             
2. 6$\mathrm{\alpha }$ and 4$\mathrm{\beta }$
3. 8$\mathrm{\alpha }$ and 24$\mathrm{\beta }$           
4. 4$\mathrm{\alpha }$ and 16$\mathrm{\beta }$

After two alpha decays and four beta(-ve) decays, the atomic number:

In a radioactive decay process, the negatively charged emitted $\beta$-particles are:

What is the respective number of $\alpha$ and $\beta$-particles emitted in the following radioactive decay?
\(X^{200}_{90}\rightarrow Y^{168}_{80}\)

A nucleus \({ }_{{n}}^{{m}} {X}\) emits one $\mathrm{}$\(\alpha\text -\text{particle}\) and two \(\beta\text- \text{particle}\) The resulting nucleus is:
1. \(^{m-}{}_n^6 Z \)
2. \(^{m-}{}_{n}^{4} X \)
3. \(^{m-4}_{n-2}Y\)
4. \(^{m-6}_{n-4} Z \)

What happens to the mass number and the atomic number of an element when it emits \(\gamma\text{-}\)radiation?

When a deuterium is bombarded on ${}_{8}O_{16}$ nucleus, an $\alpha$-particle is emitted, then the product nucleus is:

1. ${}_{7}N_{13}$ 

2. ${}_{5}B_{10}$

3. ${}_{4}B{e}_9$ 

4. ${}_{7}N_{14}$

During negative $\mathrm{\beta }$-decay: