A nucleus $_n{X^m}$emits one $\alpha $ and one $\beta $ particles. The resulting nucleus is

Actinium $231$, ${}^{231}A{c_{89}}$, emit in succession two $\beta  - $particles, four alphas, one $\beta  $ and one alpha plus several $\gamma $ rays. What is the resultant isotope

Consider the following radioactive decay process 

${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow} A_1 \stackrel{\beta^{-}}{\longrightarrow} A_2 \stackrel{\gamma}{\longrightarrow} A_3 \stackrel{\alpha}{\longrightarrow} A_4 \stackrel{B^{+}}{\longrightarrow} A_5 \stackrel{\gamma}{\longrightarrow} A_6$

The mass number and the atomic number $A _6$ are given by 

A radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles leading to ond product ${ }_{82}^{208} Pb$. $X$ is

For the given reaction, the particle $X$ is $_6{C^{11}}{ \to _5}{B^{11}} + {\beta ^ + } + X$

In the following reaction the value of $‘X’$ is $_7{N^{14}}{ + _2}H{e^4}\, \to \,X{ + _1}{H^1}$