Neon-23 decays in the following way,
`_10^23Nerarr_11^23Na+ _(-1)^0e+barv`
Find the minimum and maximum kinetic energy that the beta particle`(_(-1)^0e)`can have. The atomic masses of `^23Ne` and `^23 Na` are `22.9945u` and `22.9898u`, respectively.

Explanation

To solve the problem of finding the minimum and maximum kinetic energy of the beta particle emitted during the decay of Neon-23, we can follow these steps:

### Step 1: Identify the Reaction
The decay reaction is given as:
\[ _{10}^{23}Ne \rightarrow _{11}^{23}Na + _{-1}^{0}e + \bar{\nu} \]
This indicates that Neon-23 decays into Sodium-23, a beta particle (electron), and an antineutrino.

### Step 2: Calculate the Mass Defect
The mass defect (Δm) is calculated as the difference between the mass of the reactants and the mass of the products.

- Mass of Neon-23: \( m_{Ne} = 22.9945 \, u \)
- Mass of Sodium-23: \( m_{Na} = 22.9898 \, u \)
- Mass of the emitted beta particle (electron): \( m_e \approx 0.000548 \, u \) (this is a standard value for the mass of an electron)

The mass defect can be calculated as:
\[ \Delta m = m_{Ne} - (m_{Na} + m_e) \]
Substituting the values:
\[ \Delta m = 22.9945 \, u - (22.9898 \, u + 0.000548 \, u) = 22.9945 \, u - 22.990348 \, u = 0.004152 \, u \]

### Step 3: Convert Mass Defect to Energy
Using Einstein's mass-energy equivalence principle, we can convert the mass defect into energy using the formula:
\[ E = \Delta m \cdot c^2 \]
To convert atomic mass units (u) to MeV, we use the conversion factor \( 1 \, u \approx 931.5 \, MeV/c^2 \):
\[ E = 0.004152 \, u \cdot 931.5 \, \frac{MeV}{u} \approx 3.87 \, MeV \]

### Step 4: Determine Minimum and Maximum Kinetic Energy
- The minimum kinetic energy (K.E. min) of the beta particle can be zero (when all energy goes to the other products).
- The maximum kinetic energy (K.E. max) of the beta particle is equal to the total energy released in the decay, which is approximately \( 3.87 \, MeV \).

### Final Results
- Minimum Kinetic Energy: \( K.E._{min} = 0 \, MeV \)
- Maximum Kinetic Energy: \( K.E._{max} \approx 3.87 \, MeV \)

### Summary
The kinetic energy of the beta particle can range from 0 MeV to approximately 3.87 MeV.

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