Question

In radioactive reaction
https://cdn.mathpix.com/snip/images/f2F5c0qmJRZxp0vCFecVMy4Lt8GhR0O6j84XFXJfXug.original.fullsize.png
Successive emission of particles is

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

Answer 1

Answer: (D)

Given the radioactive reaction,
${}_Z^{A}X\to {}_{Z+1}^{A_1^A{X}_1}\to {}_{Z+2}^A{X}_2\to {}_Z^{A-4}{X}_3\to {}_{Z+1}^{A-4}{X}_4$
In the first step, the proton number increases by $1$ and hence it is ${\beta }^{-}$ decay.
In the second step, the same as first step, so ${\beta }^{-}$ decay.
In the third step, the proton number reduces by $2$ and the nucleon number reduces by $4$ , hence it is an $\alpha$ decay.
In the fourth step, it is the same as the first step, so ${\beta }^{-}$ decay.