What is the respective number of $\alpha$ and $\beta$ particles emitted in the following radioactive decay: $_{90}X^{200} \to _{80}Y^{168}$?

$A$ radioactive source is placed in a uniform electric field. If $\alpha, \beta$,and $\gamma$ particles are emitted,then $\alpha, \beta, \gamma$ particles correspond to which paths respectively?

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

In the uranium radioactive series, the initial nucleus ${}^{238}_{92}U$ decays to the final nucleus ${}^{206}_{82}Pb$. In this process, the number of $\alpha$-particles and $\beta$-particles emitted are:

In the uranium radioactive series,the initial nucleus is $_{92}U^{238}$ and the final nucleus is $_{82}Pb^{206}$. When the uranium nucleus decays to lead,the number of $\alpha$-particles emitted is $x$ and the number of $\beta$-particles emitted is $y$. Find $x$ and $y$.

The decay of a proton to a neutron is: