An atomic nucleus $_{90}Th^{232}$ emits several $\alpha$ and $\beta$-radiations and finally reduces to $_{82}Pb^{208}$. It must have emitted:

The energy spectrum of $\beta$-particles [number $N(E)$ as a function of $\beta$-energy $E$] emitted from a radioactive source is

The nucleus $_{10}^{23} Ne$ decays by $\beta^{-}$ emission. Write down the $\beta^{-}$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$

$A$ free neutron decays spontaneously into:

The total number of $\alpha$ and $\beta$ particles emitted in the nuclear reaction ${ }_{92}^{238} U \rightarrow{ }_{82}^{214} Pb$ is

$A$ common example of alpha decay is ${ }_{92}^{238} U \longrightarrow{ }_{90}^{234} Th +{ }_{2}^{4} He + Q$. (Given: ${ }_{92}^{238} U = 238.05060 \, u$,${ }_{90}^{234} Th = 234.04360 \, u$,${ }_{2}^{4} He = 4.00260 \, u$,and $1 \, u = 931.5 \, MeV/c^2$). The energy released $(Q)$ during the alpha decay of ${ }_{92}^{238} U$ is $...... \, MeV$.