Question

When a radioactive isotope, ${}_{88}{R}^{288}$ decays in series by the emission of $3\alpha -$ particles and $\beta -$ particles, the mass number of the isotope finally formed is :

• A. 83
• B. 72
• C. 216
• D. 228

Answer 1

The correct option is C 216

Decrease in mass number due to emission of $3\alpha -$ particles and $\beta -$ particles

$=3\times 4=12$

Decrease in change number in the process

$=3\times 2-1=5$

therefore $A=228-12=216$