The most penetrating radiation out of the following is

$A$ nucleus $_n{X^m}$ emits one $\alpha$ and one $\beta$ particle. The resulting nucleus is:

Radioactive substances do not emit

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.

In $\beta^{-}$ decay, a neutron transforms into a proton within the nucleus according to the equation:
$\text{neutron} \rightarrow \text{proton} + \beta^{-} + x$
In this equation, the particle represented by '$x$' is:

Three $\alpha$-particles and one $\beta$-particle decay take place in series from an isotope $_{88}Ra^{238}$. Finally,the isotope obtained will be: