Question

$_{92}U^{235}$ undergoes successive disintegrations with the end product of $_{82}Pb^{203}$.The number of $\alpha$ and $\beta$ particles emitted are

• A. $\alpha = 6, \beta = 4$
• B. $\alpha = 6, \beta = 0$
• C. $\alpha = 8, \beta = 6$
• D. $\alpha = 3, \beta = 3$

Answer 1

Answer: (C)

$\alpha = 8, \beta = 6$

Answer 2

The correct option is(C): α =8,β =6.

Let number of a particles decayed be $x$ number of $p$ particles decayed be $y$.

Then equation for the decay is given by

${ }_{92} U^{235}----->x a_{2}^{4}+y \beta_{1}^{0}+P b_{8 z}^{203}$

Equating the mass number on both sides

$235=4 x+203\,\,\,\,\,\,\,...(i)$

Equating atomic number on both sides

$92=2 x-y+82\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$x=8, y=6$

$\therefore 8 \alpha$ particles and $6 \beta$ particles are emitted in disintegration.