Question

The particles emitted during the sequential radioactive decay of ${ }^{238} U _{92}$ to ${ }^{206} Pb _{82}$ are

• A. $5\, \alpha$ and $6\, \beta$
• B. $6\, \alpha$ and $8\, \beta$
• C. $8\, \alpha$ and $4\, \beta$
• D. $8\, \alpha$ and $6\, \beta$

Answer 1

Answer: (D)

Let number of $\alpha$-particles emitted during radioactive decay $=n_{1}$
Let number of $\beta$-particles emitted during radioactive decay $=n_{2}$
${ }^{238} U _{92} \longrightarrow { }^{206} Pb _{82}+n_{1} \alpha^{+1}+n_{2} \beta^{-1}$
The $\alpha$-particle represents a helium atom. In $\alpha$-decay the atomic mass is reduced to four units.
$\therefore 238-206 =4 n_{1}$
$\Rightarrow \frac{238-206}{4} =n_{1}$
$n_{1}=8$
In $\beta$ decay the atomic number is reduced to two units.
$\therefore 92-82=2 n_{2}$
$\Rightarrow \frac{12}{2}=n_{2}$
$n_{2}=6$