Question

When a radioactive isotope ${}_{88}R{a}^{228}$ decays in series by the emission of $3$ $\alpha particles$ and beta particle, the isotope finally formed is

• A. ${}_{88}{X}^{224}$
• B. ${}_{86}{X}^{222}$
• C. ${}_{83}{X}^{216}$
• D. ${}_{83}{X}^{215}$

Answer 1

The correct option is C ${}_{83}{X}^{216}$

$\underset{–}{Given}:$

$\bullet$ Original Isotope $=$ ${}_{88}R{a}^{228}$

$\bullet$ Number of $\alpha$ particles released $=n_{\alpha }=3$

$\bullet$ Number of $\beta$ particles released $=n_{\beta }=1$

$To$ $find:$ Final Isotope ${}_Z{X}^A=$ $?$

$\underset{–}{Solution}:$

$A=228-(n_{\alpha }\times 4)$

$\therefore A=228-(3\times 4)=228-12=216$

$Z=88-(n_{\alpha }\times 2)+(n_{\beta }\times 1)$

$\therefore Z=88-(3\times 2)+(1\times 1)=83$

$\therefore$ ${}_Z{X}^A=$ ${}_{83}{X}^{216}$

$\underset{–}{Answer}:$ The isotope finally formed is ${}_{83}{X}^{216}$