The nucleus `.^(23)Ne` deacays by `beta`-emission into the nucleus `.^(23)Na`. Write down the `beta`-decay equation and determine the maximum kinetic energy of the electrons emitted. Given,`(m(._(11)^(23)Ne) =22.994466 am u` and `m (._(11)^(23)Na =22.989770 am u`. Ignore the mass of antineuttino `(bar(v))`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the beta decay equation
In beta decay, a neutron in the nucleus is converted into a proton, emitting a beta particle (electron) and an antineutrino. The equation for the beta decay of the nucleus \( ^{23}_{11}\text{Ne} \) (Neon) to \( ^{23}_{11}\text{Na} \) (Sodium) can be written as:

\[ ^{23}_{11}\text{Ne} \rightarrow ^{23}_{12}\text{Na} + e^- + \bar{\nu} \]
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