For the given reaction, the particle $X$ is $_6C^{11} \to _5B^{11} + \beta^+ + X$.

$A$ radioactive source is placed in a uniform electric field. If $\alpha, \beta$,and $\gamma$ particles are emitted,then $\alpha, \beta, \gamma$ particles correspond to which paths respectively?

$A$ radioactive nucleus ${}_{Z}^{A}X$ undergoes spontaneous decay in the sequence ${}_{Z}^{A}X \rightarrow {}_{Z-1}B \rightarrow {}_{Z-3}C \rightarrow {}_{Z-2}D$,where $Z$ is the atomic number of element $X$. The possible decay particles in the sequence are:

An atomic nucleus $_{90}^{232}Th$ emits several $\alpha$ and $\beta$ radiations and finally reduces to $_{82}^{208}Pb$. It must have emitted:

In the given radioactive decay,the number of $\alpha$ and $\beta$ particles emitted are respectively: $_{90}X^{200} \rightarrow _{80}Y^{168}$

The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} U \rightarrow{ }_{82}^{214} Pb$ is