Question

The isotope ${}_{92}^{235}U$ decays in a number of steps to an isotope of ${}_{82}^{207}Pb$. The groups of particles emitted in this process will be:

• A. $4\alpha ,7\beta$
• B. $6\alpha ,4\beta$
• C. $7\alpha ,4\beta$
• D. $10\alpha ,8\beta$

Answer 1

The correct option is C $7\alpha ,4\beta$

Decrease in atomic mass will be caused only by

$\alpha$-decay.

${}_{92}^{235}U\underrightarrow{x\alpha ,y\beta }{}_{82}^{207}Pb$

one $\alpha$-particle decreases atomic number by 2 units.

one $\alpha$-particle decreases atomic number by 4 units.

one $\beta$-particle increases atomic number by 1 units.

No. of $\alpha$-particles emitted = $\frac{235-207}{4}$

$\Rightarrow 7$

No. of $\beta$-particles emitted (y) = 7(2)-(92-82)

$\Rightarrow 4$