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 products of the neutron. Therefore, considering the decay of a neutron as a tro-body dcay process, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process i.e., `n rarr p + e^(-)+overset(-)v_(e )`, around `1930`, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino `(overset(-)V_(e ))` to be massless and possessing negligible energy, and neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is `0.8xx10^(6)eV`. The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino has a mass of `3eV//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?

Options

A. `0 leK le0.8xx10^(6)eV`
B. `3.0 eV le K le 0.8xx10^(6) eV`
C. `3.0 eV le K lt 0.8xx10^(6)eV`
D. `0 le K lt 0.8xx10^(6) eV`

Explanation

To solve the problem regarding the kinetic energy of the electron in the beta decay process when the anti-neutrino has a mass, we can follow these steps:

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

### Step 2: Conservation of Energy and Momentum
In the original scenario where the anti-neutrino is considered massless, the maximum kinetic energy of the electron was calculated to be:
\[ K_{max} = 0.8 \times 10^6 \, \text{eV} \]

Now, if we assume the anti-neutrino has a mass of \( 3 \, \text{eV}/c^2 \), we need to account for the energy that the anti-neutrino will take away from the total energy available in the decay.

### Step 3: Total Energy Consideration
The total energy available in the decay process when the neutron is at rest is given by the mass-energy equivalence:
\[ E_{total} = m_n c^2 \]
where \( m_n \) is the mass of the neutron.

### Step 4: Energy Distribution
When the anti-neutrino has mass, it will carry away some energy. The energy carried by the anti-neutrino can be expressed as:
\[ E_{\overline{\nu}_e} = m_{\overline{\nu}_e} c^2 = 3 \, \text{eV} \]

Thus, the total energy available for the electron and proton will be:
\[ E_{available} = E_{total} - E_{\overline{\nu}_e} \]

### Step 5: Maximum Kinetic Energy of the Electron
Since the maximum kinetic energy of the electron occurs when the anti-neutrino takes the least energy (which is its rest mass energy), we can express the maximum kinetic energy of the electron as:
\[ K_{max}' = K_{max} - E_{\overline{\nu}_e} \]
Substituting the values:
\[ K_{max}' = 0.8 \times 10^6 \, \text{eV} - 3 \, \text{eV} \]
\[ K_{max}' \approx 0.8 \times 10^6 \, \text{eV} \]

### Step 6: Minimum Kinetic Energy of the Electron
The minimum kinetic energy of the electron occurs when the anti-neutrino takes all the energy, which theoretically would be:
\[ K_{min} = 0 \, \text{eV} \]

### Step 7: Conclusion
Thus, the range of the kinetic energy \( K \) of the electron when the anti-neutrino has a mass of \( 3 \, \text{eV}/c^2 \) is:
\[ 0 \leq K < 0.8 \times 10^6 \, \text{eV} \]

### Final Answer
The range of the kinetic energy \( K \) of the electron is:
\[ 0 \leq K < 0.8 \times 10^6 \, \text{eV} \]

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