Question

A radioactive element X disintegrates successively as under:
$X\stackrel{\alpha }{\to }{X}_1\stackrel{\beta }{\to }{X}_2\stackrel{\alpha }{\to }{X}_3\stackrel{\gamma }{\to }{X}_4$
If the atomic number and atomic mass. Number of X are $72$ and $180$. What are the corresponding values for $X_4$.

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

Answer 1

The correct option is B $69,172$

In $\alpha$ decay the mass number decreases by 4 and atomic number decreases by 2.

In $\beta$ decay the mass number remains constant but atomic number increases by 1.

In $\gamma$ decay the mass number and atomic number remain constant.

In $X\stackrel{\alpha }{\to }{X}_1\stackrel{\beta }{\to }{X}_2\stackrel{\alpha }{\to }{X}_3\stackrel{\gamma }{\to }{X}_4$

Given: ${}_{72}^{180}X$

Now

$X_1$ will be ${}_{72-2}^{180-4}{X}_1{=}_{70}^{176}{X}_1$

$X_2$ will be ${}_{70+1}^{176}{X}_2{=}_{71}^{176}{X}_2$

$X_3$ will be ${}_{71-2}^{176-4}{X}_3{=}_{69}^{172}{X}_3$

$X_4$ will be ${}_{69\pm 0}^{172\pm 0}{X}_4{=}_{69}^{172}{X}_4$

Hence the atomic mass$=69$ and mass number $=172$ for $X_4$.

Therefore the correct option is (B).