Question

In the radioactive decay process of uranium the initial nuclide is ${}_{92}{U}^{238}$ and the final nuclide is ${}_{82}{Pb}^{206}$. When uranium nucleus decays to lead, then the number of $\alpha$ -particles and $\beta$ -particles emitted will respectively be :

• A. $8,6$
• B. $8,4$
• C. $6,8$
• D. $4,8$

Answer 1

The correct option is A $8,6$

In alpha decay, mass number $A$ and atomic number $Z$ reducing by $4$ and $2$ respectively. In beta decay, mass number unchanged but atomic number increasing by $1.$
The number of emitted alpha particles $=\frac{238-206}{4}=8$
For 8 alpha particles the atomic number $A$ reduce to $92-(2\times 8)=92-16=76$
So for $A=82$ we need to increase $6$ that is given by $6$ beta particles emission.