During a negative beta decay,

$A$ nucleus with $Z = 92$ emits $\alpha, \alpha, \beta^-, \beta^-, \alpha, \alpha, \alpha, \alpha, \beta^-, \beta^-, \alpha, \beta^+, \beta^+, \alpha$ particles in sequence. What is the $Z$ of the resulting nucleus?

In the nuclear reaction ${}_{92}^{235}U$ decaying to ${}_{91}^{231}Pa$,what are the particles emitted?

In a radioactive decay chain,${ }_{90}^{232} Th$ nucleus decays to ${ }_{82}^{212} Pb$ nucleus. Let $N_{\alpha}$ and $N_{\beta}$ be the number of $\alpha$ and $\beta^{-}$ particles,respectively,emitted in this decay process. Which of the following statements is (are) true?
$(A)$ $N_{\alpha}=5$
$(B)$ $N_{\alpha}=6$
$(C)$ $N_{\beta}=2$
$(D)$ $N_{\beta}=4$

$1 \text{ Curie}$ is equal to

The radionuclide $^{11} C$ decays according to
$_{6}^{11} C \rightarrow_{5}^{11} B + e^{+} + \nu: \quad T_{1/2} = 20.3 \; min$
The maximum energy of the emitted positron is $0.960 \; MeV$. Given the mass values:
$m(_{6}^{11} C) = 11.011434 \; u$ and $m(_{5}^{11} B) = 11.009305 \; u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.