A nucleus `""_(10)Ne^(23)` undergoes `beta`- decay and becomes `""_(11)Na^(23)` . Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and antineutrino carry negligible kinetic energy. Given mass of `""_(10)Ne^(23)` = 22.994466 u and mass of `""_(11)Na^(23)` = 22.989770 m

Answer

To solve the problem of calculating the maximum kinetic energy of electrons emitted during the beta decay of the nucleus \( _{10}^{23}\text{Ne} \) to \( _{11}^{23}\text{Na} \), we can follow these steps:

### Step 1: Understand the decay process
In beta decay, a neutron in the nucleus is converted into a proton, emitting a beta particle (electron) and an antineutrino. The reaction can be written as:
\[ _{10}^{23}\text{Ne} \rightarrow _{11}^{23}\text{Na} + \beta^- + \bar{\nu} \]
Here, \( \beta^- \) represents the emitted electron and \( \bar{\nu} \) is the antineutrino.

### Step 2: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the mass of the parent nucleus and the mass of the daughter nucleus. We can denote the mass of Neon as \( m_{\text{Ne}} \) and the mass of Sodium as \( m_{\text{Na}} \).

Given:
- Mass of \( _{10}^{23}\text{Ne} = 22.994466 \, \text{u} \)
- Mass of \( _{11}^{23}\text{Na} = 22.989770 \, \text{u} \)

The mass defect can be calculated as:
\[ \Delta m = m_{\text{Ne}} - m_{\text{Na}} = 22.994466 \, \text{u} - 22.989770 \, \text{u} \]
\[ \Delta m = 0.004696 \, \text{u} \]

### Step 3: Convert mass defect to energy
Using Einstein's mass-energy equivalence principle, we can convert the mass defect into energy. The energy equivalent of 1 atomic mass unit (u) is approximately 931.5 MeV. Therefore, the energy released (\( Q \)) in the decay can be calculated as:
\[ Q = \Delta m \times 931.5 \, \text{MeV/u} \]
Substituting the value of \( \Delta m \):
\[ Q = 0.004696 \, \text{u} \times 931.5 \, \text{MeV/u} \]
\[ Q \approx 4.37 \, \text{MeV} \]

### Step 4: Determine the maximum kinetic energy of the emitted electron
In beta decay, the total energy released is shared between the emitted electron and the antineutrino. Assuming the antineutrino carries negligible kinetic energy, the maximum kinetic energy of the emitted electron (\( K_{\text{max}} \)) is approximately equal to the total energy released:
\[ K_{\text{max}} \approx Q \approx 4.37 \, \text{MeV} \]

### Final Answer
The maximum kinetic energy of the electrons emitted during the beta decay of \( _{10}^{23}\text{Ne} \) to \( _{11}^{23}\text{Na} \) is approximately **4.37 MeV**.

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