**Problem (IIT JEE 2012):**

*PARAGRAPH: *
The $\beta$-decay process, discovered around 1900, is basically the decay of a neutron $(n)$. In the laboratory, a proton $(p)$ and an electron $(e^-)$ are observed as the decay product of the neutron. Therefore, considering the decay of a neutron as a two body decay process, it was predicted theoretically that the kinetic energy of the electron should be constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., $n\rightarrow p+e^- +\bar{\nu}_e$, around 1930, Pauli explained the observed electron energy spectrum. Assuming the antineutrino ($\bar{\nu}_e$) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is ${0.8\times{10}^{6}}\;\mathrm{eV}$. The kinetic energy carried by the proton is only the recoil energy.

**Question 1: ** What is the maximum energy of the anti-neutrino?

- Zero.
- Much less than ${0.8\times{10}^{6}}\;\mathrm{eV}$.
- Nearly ${0.8\times{10}^{6}}\;\mathrm{eV}$.
- Much larger than ${0.8\times{10}^{6}}\;\mathrm{eV}$.

**Solution:**
The energy released in $\beta$-decay is given by,
\begin{align}
Q&=\left[m_n-(m_p+m_e+m_{\bar{\nu}_e})\right]c^2\nonumber\\
&=\left[m_n-(m_p+m_e)\right]c^2 \quad (\because m_{\bar{\nu}_e}=0). \nonumber
\end{align}
This energy is in the form of kinetic energy of released particles i.e.,
\begin{align}
Q&=K_p+K_e+K_{\bar{\nu}_e} \approx K_e+K_{\bar{\nu}_e}\qquad (\because K_p\approx 0).\nonumber
\end{align}
The conservation of linear momentum gives,
\begin{align}
\vec{p}_p+\vec{p}_e+\vec{p}_{\bar{\nu}_e}=0.\nonumber
\end{align}
The $K_e$ is maximum when $K_{\bar{\nu}_e}=0$ (and hence $p_{\bar{\nu}_e}=0$). This is possible in three body decay without violating momentum conservation. Thus,
\begin{align}
Q=K_{e}^\text{max}={0.8\times{10}^{6}}\;\mathrm{eV}.\nonumber
\end{align}
Similarly, $K_{\bar{\nu}_e}$ is maximum when $K_{e}=0$ i.e.,
\begin{align}
K_{\bar{\nu}_e}^\text{max}=Q={0.8\times{10}^{6}}\;\mathrm{eV}.\nonumber
\end{align}

**Question 2:**
If the anti-neutrino had a mass of ${3}\;\mathrm{eV/c^2}$ (where $c$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy, $K$, of the electron?

- $0 \leq K \leq {0.8\times{10}^{6}}\;\mathrm{eV}$
- ${3.0}\;\mathrm{eV} \leq K \leq {0.8\times{10}^{6}}\;\mathrm{eV}$
- ${3.0}\;\mathrm{eV} \leq K < {0.8\times{10}^{6}}\;\mathrm{eV}$
- $0 \leq K < {0.8\times{10}^{6}}\;\mathrm{eV}$

**Solution: **
If $m_{\bar{\nu}_e}={3}\;\mathrm{eV/c^2}$, then,
\begin{align}
Q={0.8\times{10}^{6}}\;\mathrm{eV}-{3}\;\mathrm{eV} < {0.8\times{10}^{6}}\;\mathrm{eV}.\nonumber
\end{align}
Thus, $K_{\bar{\nu}_e}^\text{max}=Q < {0.8\times{10}^{6}}\;\mathrm{eV}$. The $K_{\bar{\nu}_e}^\text{min}$ is related to $p_{\bar{\nu}_e}^\text{min}$ which can be zero in three body decay making $K_{\bar{\nu}_e}^\text{min}=0$.

**Problem (IIT JEE 2001): **
The electron emitted in beta radiation originates from

- inner orbits of atom.
- free electrons existing in nuclei.
- decay of a neutron in a nucleus.
- photon escaping from the nucleus.

**Solution: **
Beta rays are the electrons emitted by the nucleus when a neutron is converted to a proton $\sideset{_0^1}{}{n}\to \sideset{_1^1}{}{p}+\sideset{_{-1}^{\;\; 0}}{}{e}$.

**Problem (IIT JEE 1999) **
Which of the following is a correct statement?

- Beta rays are same as cathode rays.
- Gamma rays are high energy neutrons.
- Alpha particles are singly ionized helium atoms.
- Protons and neutrons have exactly the same mass.

The $\beta$ particles are electrons emitted by the nucleus whenever a neutron is converted to a proton. The gamma rays are electromagnetic radiations. Alpha particle is doubly ionized helium atom. The rest mass of a neutron is slightly greater than that of a proton.