Question

One form of nuclear radiation, beta decay, occurs when a neutron changes into a proton, an electron, and a neutral particle called a neutrino: $n\to p^{+}+e^{-}+v$where $n$is the symbol for a neutrino. When this change happens to a neutron within the nucleus of an atom, the proton remains behind in the nucleus while the electron and neutrino are ejected from the nucleus. The ejected electron is called a beta particle. One nucleus that exhibits beta decay is the isotope of hydrogen $3H$, called tritium, whose nucleus consists of one proton (making it hydrogen) and two neutrons (giving tritium an atomic mass $m=3u$). Tritium is radioactive, and it decays to helium: ${}^{3}H\to {}^{3}He+e^{-}+n$

a. Is charge conserved in the beta decay process? Explain.

b. Why is the final product a helium atom? Explain.

c. The nuclei of both ${}^{3}H\mathrm{and}{}^{\mathit{3}}He$ have radii of $1.5\times {10}^{-15}m$. With what minimum speed must the electron be ejected if it is to escape from the nucleus and not fall back?

Short answer

a. The total charge remains constant and the charge is conserved .

b. The nucleus converts to He nucleus when the quantity of proton in the nucleus reaches two.

c. In order for the electron to escape and not fall back, it must be ejected at a speed of $8.2\times {10}^{8}m/s$

Step-by-step solution

Given information and theory used

Given :

Neutrino : $n\to p^{+}+e^{-}+v$

Helium: ${}^{3}H\to {}^{3}He+e^{-}+n$

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

Theory used :

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

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

Determining if charge is conserved in the beta decay process 

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.

Determining why is the final product a helium atom 

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.

Calculating the minimum speed required

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

$$\begin{gathered} \frac{1}{4\pi {\epsilon }_0}\frac{Z{e}^2}{r}=\frac{1}{2}m{v}^2 \\ \Rightarrow v=\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{2Z{e}^2}{mr}} \end{gathered}$$

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

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