Question

${}_{92}^{235}\text{U}$ atom disintegrates to ${}_{82}^{207}\text{Pb}$ with a half-life of ${10}^9$ years. In the process it emits 7 alpha particles and n ${\beta }^{-}$ particles. Here n is -

• A. 7
• B. 3
• C. 4
• D. 14

Answer 1

The correct option is C 4

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 $=207-235=-28$ (only due to $\alpha -$emission) ,

hence , number of $\alpha -$ particles emitted $=28/4=7$ ,

now , due to emission of 7 $\alpha -$ particles , atomic number will be $=92-7\times 2=78$ , but the final atomic number is 82 therefore difference in atomic number $=82-78=4$ .

As emission of one $\beta -$ particle increases the atomic number by one ,hence atomic number will be increased by 6 (upto 82) , by the emission of 4 $\beta -$ particles .