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.
If the - neutrono had a mass of `3 eV// c^(2)` (where c is the speed of light ) insend of zero mass , what should be the range of the kinectic energy `K.` of the electron ?

Explanation

To solve the problem regarding the kinetic energy of the electron produced in the beta decay of a neutron, we can follow these steps:

### Step 1: Understand the decay process
The beta decay process can be represented as:
\[ n \rightarrow p + e^- + \bar{\nu}_e \]
where \( n \) is the neutron, \( p \) is the proton, \( e^- \) is the electron, and \( \bar{\nu}_e \) is the anti-neutrino. Initially, it was thought to be a two-body decay, but it was later understood to be a three-body decay due to the presence of the anti-neutrino.

### Step 2: Conservation of Energy and Momentum
In a three-body decay, both energy and momentum must be conserved. The total energy before decay (mass-energy of the neutron) must equal the total energy after decay (mass-energy of the proton, electron, and anti-neutrino).

### Step 3: Maximum Kinetic Energy Calculation
When the anti-neutrino is assumed to be massless and at rest, the maximum kinetic energy of the electron was found to be:
\[ K_{max} = 0.8 \times 10^6 \text{ eV} \]
This means that the electron can take up to this amount of kinetic energy if the anti-neutrino carries negligible energy.

### Step 4: Considering the Anti-Neutrino Mass
If we assume that the anti-neutrino has a small mass and carries some kinetic energy, the kinetic energy of the electron will be less than the maximum value calculated earlier. The energy conservation can be expressed as:
\[ K_e + K_p + K_{\bar{\nu}} = Q \]
where \( K_e \) is the kinetic energy of the electron, \( K_p \) is the recoil energy of the proton, \( K_{\bar{\nu}} \) is the kinetic energy of the anti-neutrino, and \( Q \) is the available energy from the decay.

### Step 5: Establish the Range of Kinetic Energy
Since the anti-neutrino now has mass and carries some kinetic energy, the kinetic energy of the electron must satisfy:
\[ 0 \leq K_e < K_{max} \]
Thus, the range of kinetic energy \( K_e \) of the electron can be expressed as:
\[ 0 \leq K_e < 0.8 \times 10^6 \text{ eV} \]

### 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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