Question

When a radioactive isotope ${}_{88}R{a}^{228}$ decays in series by the emission of three $\alpha$-particles and a $\beta$ particle the isotope finally formed is :

• A. ${}_{84}{X}^{220}$
• B. ${}_{86}{X}^{222}$
• C. ${}_{83}{X}^{216}$
• D. ${}_{83}{X}^{215}$

Answer 1

Answer: (C)

${}_{83}{X}^{216}$

In each alpha decay, mass number of the parent nucleus decreases by $4$ units while atomic number decreases by $2$ units. In each beta decay, atomic number increases by $1$ unit but mass number remains the same.
So, mass number of the isotope formed $A=228-3\times 4=216$
Atomic number $Z=88-3\times 2+1=83$
So correct answer is option C.