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

The Correct Option is C

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.