Question

In the nuclear reaction ${}_{92}{U}^{238}\to {}_{82}P{b}^{206}$ the number of $\alpha$ and $\beta$-particles emitted is

• A. $7\alpha ,5\beta$
• B. $6\alpha ,4\beta$
• C. $4\alpha ,3\beta$
• D. $8\alpha ,6\beta$

Answer 1

The correct option is D $8\alpha ,6\beta$

We know that due to one $\alpha$-particle emission , mass number of the nucleus decreases by 4 and atomic number decreases by 2 and due to one $\beta$-particle emission, mass number of the nucleus remains unchanged and atomic number increases by 1.

In the given reaction,

change in mass number $=206-238=-32$ (only due to $\alpha -$emission) ,

hence, number of $\alpha -$ particles emitted $=32/4=8$ ,

now, due to emission of 8 $\alpha -$ particles , atomic number will be $=92-8\times 2=76$ , but the final atomic number is 82 therefore difference in atomic number $=82-76=6$.

As emission of one $\beta -$ particle increases the atomic number by one hence atomic number will be increased by 6, by the emission of 6 $\beta -$ particles.