The nucleus $_{90}Th^{228}$ transforms into $_{83}Bi^{212}$. How many $\alpha$-particles and $\beta$-particles are emitted during this process?

For the given reaction, the particle $X$ is $_6C^{11} \to _5B^{11} + \beta^+ + X$.

$A$ 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 $...... \, MeV$.

The electron emitted in beta radiation originates from

At a specific instant,the emission of a radioactive compound is deflected in a magnetic field. The compound can emit:
$(i)$ Electrons
$(ii)$ Protons
$(iii)$ $He^{2+}$
$(iv)$ Neutrons
The emission at that instant can be:

If a nucleus emits an $e^-$ particle,then its neutron-to-proton ratio $[n/p]$ will .....