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

$A$ radioactive element $A$ decays into radioactive element $C$ by the following processes in succession.
$A \rightarrow B + {}_{2}^{4}He$
$B \rightarrow C + 2e^{-}$
Then elements

$A$ certain radioactive material can undergo three different types of decay,each with a different decay constant $\lambda_1$,$\lambda_2$,and $\lambda_3$. Then the effective decay constant is:

$A$ radioactive nucleus undergoes a series of decays according to the scheme
$A \xrightarrow{\alpha} A_1 \xrightarrow{\beta} A_2 \xrightarrow{\alpha} A_3 \xrightarrow{\gamma} A_4$
If the mass number and atomic number of $A$ are $180$ and $72$,respectively,then what are these numbers for $A_4$?

$_6^{12}C$ absorbs an energetic neutron and emits a beta particle. The resulting nucleus is

$A$ radioactive nucleus (initial mass number $A$ and atomic number $Z$) emits $3$ $\alpha$-particles and $2$ positrons. The ratio of the number of neutrons to that of protons in the final nucleus will be: