Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The beta decay process is given by
$${\rm n}\rightarrow {\rm p^+}+{\rm e^-}+\nu$$
The neutron is neutral, the proton has a charge of $+e$, the electron has a charge of $-e$, and the neutrino has no charge.
Thus, the left side of the formula above has zero net charge and the right side has a net charge of zero as well.
$$+e+(-e)+0=0$$
Therefore, yes, the charge is conserved in the beta decay process.
$$\color{blue}{\bf [b]}$$
The change of the number of the neutrons of an elements makes many isotopes of it but the change in the number of protons creates a new different eelment.
Tritium has one proton and two neutrons. So when one of the neutrons decay to [proton+electron+neutrino] where the electron and the neutrono ejected away from the atom, the necleus now contians two protons and one neutron.
This is now a helium isotob atom with mass of 3 u.
$$\color{blue}{\bf [c]}$$
We need to find the initial speed of the electron when it is ejected away from the helium atom so it reaches a distance of $r\rightarrow\infty$ and then stops.
Using the conservation of energy,
$$E_i=E_f$$
$$K_i+U_i=K_f+U_f$$
where $K_f=0$ since the electron finally stops, and $U_f=0$ since $r$ approches $\infty$,
$$\frac{1}{2}m_ev_e^2+\dfrac{kq_eq_{He}}{r_{He}}=0+0$$
where $q_e=-e$, and $q_{He}=+2e$
$$\frac{1}{2}m_ev_e^2+\dfrac{-2ke^2 }{r_{He}}=0 $$
Hence,
$$ v_e =\sqrt{\dfrac{4ke^2 }{r_{He}m_e } }$$
Plugging the known;
$$ v_e =\sqrt{\dfrac{4(9\times 10^9)(1.6\times 10^{-19})^2 }{(1.5\times 10^{-15})(9.11\times 10^{-31}) } }$$
$$v_e=\color{red}{\bf 8.21\times 10^8}\;\rm m/s$$
This speed is greater than the speed of light, so we had to include the relativistic analysis to get the perfect result. But this answer is enough for this chapter.
