The nucleus $_{48}^{115}Cd$ after two successive ${\beta ^ - }$ decays will give:

$A$ 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:

Actinium $231$,${}^{231}Ac_{89}$,emits in succession two $\beta^-$ particles,four $\alpha$ particles,one $\beta^-$ particle,and one $\alpha$ particle,plus several $\gamma$ rays. What is the resultant isotope?

Which of the following is deflected by an electric field?

$A$ positron is emitted from ${}^{23}Na_{11}$. The ratio of the atomic mass and atomic number of the resulting nuclide 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 $A_4$?