Question

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

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

Answer 1

Answer: (D)

$70,172$

We know that atomic number of the parent nucleus decreases by $2$ and mass number decreases by $4$ in each alpha decay whereas atomic number of the parent nucleus increases by $1$ while mass number remains the same in each beta minus decay.
So atomic number of $X_4$, $Z=72+2\times 1-2\times 2=70$
So mass number of $X_4$, $A=180-2\times 4=172$
Correct option is D.