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Given :

Neutrino : $n鈫�p^{+}+e^{-}+v$

Helium: ${}^{3}H鈫�{}^{3}He+e^{-}+n$

${}^{3}H\mathrm{and}{}^{\mathit{3}}He$have radii of : $1.5脳{10}^{-15}m$

Theory used :

$U=-\frac{1}{4蟺{蔚}_0}\frac{Z{e}^2}{r}$is the electrical potential energy of an electron.

$E_k=\frac{1}{2}m{v}^2$gives the kinetic energy.

One neutron is transformed into one proton, one electron, and one neutrino after beta decay. The charge of the proton and electron is equal and opposite, but the charge of the neutrino is less. As a result, the total charge is nil. As a result, the charge is preserved.

Total charge of ${}^{3}H$before decay is $+e$, total charge of ${}^{3}He$nucleus after beta decay is $+2e$, and electron charge is $-e$

Therefore total charge is

$(+2e)+(-e)=+e.$

As a result, the total charge before and after decay is the same. Henceforth, the charge is Conserved.

The atomic number, or the number of protons in the nucleus, determines the element. There was one proton and two neutrons before decay. One neutron decays into a proton by ejecting one electron.

As a result, the nucleus' proton count is reduced to two. So, the nucleus is renamed $He$nucleus.

The attracting force is now represented by the negative sign in the potential energy. The magnitudes of the energy will now be equal.

As a result, we must use

$$\begin{gathered} \frac{1}{4蟺{蔚}_0}\frac{Z{e}^2}{r}=\frac{1}{2}m{v}^2 \\ 鈬�v=\sqrt{\frac{1}{4蟺{蔚}_0}\frac{2Z{e}^2}{mr}} \end{gathered}$$

Now, in the previous equation, substitute all the parameters in the right-hand side to compute the velocity

localid="1650965995149" role="math" $$\begin{gathered} v=\sqrt{(8.9脳{10}^9\text{m/F})\frac{2\left(2\right)(1.6x{10}^{-19})C^2}{\left(9.11x{10}^{-31}\text{kg}\right)(1.5x{10}^{15}\text{m})}} \\ =\boxed{\mathbf{8}\mathbf{.}\mathbf{2}\mathbf{脳}{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathbf{\text{m/s}}\mathbf{}} \end{gathered}$$