Question

A radioactive element $X$ distintegrates successively
$X\stackrel{{\beta }^{-}}{\to }{X}_1\stackrel{\alpha }{\to }{X}_2\stackrel{{\beta }^{-}}{\to }{X}_3\stackrel{\alpha }{\to }{X}_4$
If atomic number and atomic mass number of $X$ are respectively $72$ and $180$, what are the corresponding values for $X_4$?

• A. $69,172$
• B. $69,176$
• C. $71,176$
• D. $70,172$

Answer 1

The correct option is D $70,172$

The disintegration can be shown as below.
${}_{72}{X}_{180}\frac{-2\overline{\beta }}{-2\alpha }{X}_4$
Loss in atomic number $=72+2-4=70$
Loss in mass number $=180-8=172$
So, final nucleus is ${}_{70}{X}_4^{172}$.