Question

Element ${}_Z{M}^A$ emits one $\alpha$ (alpha) particle followed by three $\beta$(beta) particles. Among the following, the daughter element is,

• A. ${}_{Z-2}{M}^{A-4}$
  
• B. ${}_{Z-2}{M}^A$
  
• C. ${}_{Z+1}{M}^{A-4}$
  
• D. ${}_{Z+2}{M}^{A-4}$
  

Answer 1

The correct option is C ${}_{Z+1}{M}^{A-4}$


For $\alpha -$emission.
${}_Z{M}^A{\to }_{Z-2}{M}^{A-4}$ (one $\alpha$ particle emission)
For two $\beta -$particle emission, mass number remains the same but atomic number increases by 1, for each emission.
${}_{Z-2}{M}^{A-4}{\to }_{Z-2+3}{M}^{A-4}$ (two $\beta$ particle emission)
So, the final element will be, ${}_{Z+1}{M}^{A-4}$

Hence, $\left(C\right)$ is the correct answer.