Question

A radioactive nucleus ${}_Z^{A}X$ undergoes spontaneous decay in the sequence

${}_Z^{A}X{\to }_{z-1}B{\to }_{z-3}C{\to }_{z-2}D$

where $Z$ is the atomic number of element $X.$ The possible decay particles in the sequence are

• A. ${\beta }^{-},\alpha ,{\beta }^{+}$
• B. $\alpha ,{\beta }^{-},{\beta }^{+}$
• C. $\alpha ,{\beta }^{+},{\beta }^{-}$
• D. ${\beta }^{+},\alpha ,{\beta }^{-}$

Answer 1

The correct option is D ${\beta }^{+},\alpha ,{\beta }^{-}$

We know that,
On ${\beta }^{+}$ decay, atomic number decreases by $1$

On ${\beta }^{-1}$ decay, atomic number increases by $1$

On $\alpha$ decay, atomic number decreases by $2$

${}_Z^{A}X\stackrel{{\beta }^{+}\text{decay}}{\to }\underset{Z-1}{}B\stackrel{\alpha \text{decay}}{\to }\underset{Z-3}{}C\stackrel{{\beta }^{-}\text{decay}}{\to }\underset{Z-2}{}D$

Thus, correct order of decay are ${\beta }^{+},\alpha ,{\beta }^{-}$.

Hence, option $\left(\text{D}\right)$ is correct.