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^(bar)` are observed as the decay product of neutron. Therefore considering the decay of neutron as a two- body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant . But experimentally , it was observed that the electron kinetic energy has continuous spectrum Considering a three- body decay process , i.e.
` n rarr p + e^(bar) + bar nu _(e) , ` around `1930` , Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(bar nu_(e))` to be massaless and possessing negligible energy , and the neutrino to be at rest , momentum and energy conservation principle are applied. From this calculation , the maximum kinetic energy of the electron is `0.8 xx 10^(6) eV` The kinetic energy carried by the proton is only the recoil energy.
What is the maximum energy of the anti-neutrino ?

Options

A. The nucleus `._3^6Li` can emit an alpha particle
B. The nucleus `._(84)^(120) P0` cam emit a proton.
C. Deuteron and alpha particle can undergo complete fusion.
D. The nuclei `._(30)^(70) Zn` and `._(34)^(82) Se` can undergo complete fusion.

Explanation

To find the maximum energy of the anti-neutrino in the beta decay process, we can follow these steps:

### Step 1: Understand the decay process
The beta decay process can be expressed as:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
where:
- \( n \) is the neutron,
- \( p \) is the proton,
- \( e^- \) is the electron,
- \( \bar{\nu}_e \) is the anti-neutrino.

### Step 2: Apply conservation of energy
In a three-body decay, the total energy before the decay must equal the total energy after the decay. Initially, the neutron has a certain rest mass energy, which is given by:
\[ E_n = m_n c^2 \]
After decay, the total energy is the sum of the energies of the decay products:
\[ E_{\text{total}} = E_p + E_e + E_{\bar{\nu}_e} \]
where \( E_p \) is the energy of the proton, \( E_e \) is the energy of the electron, and \( E_{\bar{\nu}_e} \) is the energy of the anti-neutrino.

### Step 3: Consider the maximum kinetic energy of the electron
From the problem, we know that the maximum kinetic energy of the electron is:
\[ KE_e^{\text{max}} = 0.8 \times 10^6 \, \text{eV} \]

### Step 4: Analyze the energy of the proton
The kinetic energy of the proton is primarily its recoil energy, which is much smaller than the kinetic energy of the electron. For maximum energy transfer to the anti-neutrino, we can assume that the electron has zero kinetic energy:
\[ KE_p \approx 0 \]

### Step 5: Write the energy conservation equation
If we assume that the proton's kinetic energy is negligible, we can express the conservation of energy as:
\[ m_n c^2 = KE_p + KE_e^{\text{max}} + E_{\bar{\nu}_e} \]
Since \( KE_p \) is negligible, we can simplify this to:
\[ m_n c^2 \approx KE_e^{\text{max}} + E_{\bar{\nu}_e} \]

### Step 6: Solve for the maximum energy of the anti-neutrino
Rearranging the equation gives us:
\[ E_{\bar{\nu}_e} = m_n c^2 - KE_e^{\text{max}} \]
Since we are interested in the maximum energy of the anti-neutrino when the electron has maximum kinetic energy:
\[ E_{\bar{\nu}_e}^{\text{max}} = KE_e^{\text{max}} \]
Thus, we can conclude that:
\[ E_{\bar{\nu}_e}^{\text{max}} = 0.8 \times 10^6 \, \text{eV} \]

### Final Answer
The maximum energy of the anti-neutrino is:
\[ E_{\bar{\nu}_e}^{\text{max}} = 0.8 \times 10^6 \, \text{eV} \]

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