$A$ radioactive element ${}_{92}^{242}X$ emits two $\alpha$-particles,one electron,and two positrons. The product nucleus is represented by ${}_{P}^{234}Y$. The value of $P$ is $..................$

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?

In the uranium radioactive series,the initial nucleus is ${ }_{92}^{238} U$ and the final nucleus is ${ }_{82}^{206} Pb$. When the uranium nucleus decays into lead,the number of $\alpha$-particles and $\beta$-particles emitted are

$A$ radioactive nucleus emits $4 \alpha$ particles and $7 \beta$ particles in succession. The ratio of the number of neutrons to that of protons in the final nucleus is $[A = \text{mass number}, Z = \text{atomic number}]$

$A$ nucleus of an element ${}_{84}X^{202}$ emits an $\alpha$-particle first,a $\beta$-particle next,and then a gamma photon. The final nucleus formed has an atomic number:

$A$ nucleus $X$ emits a beta particle to produce a nucleus $Y$. If their atomic masses are $M_{x}$ and $M_{y}$ respectively,the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light):