Question

A radioactive nucleus undergoes a series of decays according to the scheme
$A\stackrel{\alpha }{\to }{A}_1\stackrel{\beta }{\to }{A}_2\stackrel{\alpha }{\to }{A}_3\stackrel{\gamma }{\to }{A}_4$
If the mass number and atomic number of A are $180$ and $72,$ respectively, then what are these number for $A_4$?

• A. $172$ and $69$
• B. $174$ and $70$
• C. $176$ and $69$
• D. $176$ and $70$

Answer 1

The correct option is A $172$ and $69$

$\alpha$ decay decreases mass number by $4$ and atomic number by $2$.
$\alpha =2n+2p$
$\beta$ decay increases the atomic number by $1$.
So, according to the reaction:
$\Delta M=-4+0-4=-8$
$\Delta Z=-2+1-2=-3$
So, $M^{\prime }=180-8=172$
$Z^{\prime }=72-3=69$