A nucleus with $Z = 92$ emits the following in a sequence: $\alpha ,\,{\beta ^ - },\,{\beta ^ - },\,\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },\,{\beta ^ - },\alpha ,\,{\beta ^ + },\,{\beta ^ + },\,\alpha $. The $Z$ of the resulting nucleus is
$74$
$76$
$78$
$82$
In the given nuclear reaction $A, B, C, D, E$ represents
$_{92}{U^{238}}{\xrightarrow{\alpha }_B}T{h^A}{\xrightarrow{\beta }_D}P{a^C}{\xrightarrow{E}_{92}}{U^{234}}$
The isotope ${ }_5^{12} \mathrm{~B}$ having a mass $12.014 \mathrm{u}$ undergoes $\beta$-decay to ${ }_6^{12} \mathrm{C} .{ }_6^{12 .}$ has an excited state of the nucleus $\left({ }_6^{12} \mathrm{C}^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} \mathrm{~F}$ decays to ${ }_6^{12} \mathrm{C}^*$, the maximum kinetic energy of the $\beta$-particle in units of $\mathrm{MeV}$ is ( $1 \mathrm{u}=931.5 \mathrm{MeV} / c^2$, where $c$ is the speed of light in vacuum).
The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} \mathrm{U} \rightarrow{ }_{82}^{214} \mathrm{~Pb}$ is
Originally the radioactive beta decay was thought as a decay of a nucleus with the emission of electrons only (Case $I$) . However, in addition to the electron, another (nearly) massless and electrically neutral particle is also emitted (Case $II$). Based on the figure below, which of the following is correct?
A radioactive nucleus $_{92}{X^{235}}$decays to $_{91}{Y^{231}}$. Which of the following particles are emitted