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(._(10)^(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 need to follow these steps:

### Step 1: Write the beta decay equation
The beta decay of the nucleus \( ^{23}_{10}\text{Ne} \) (Neon) can be represented as:
\[ ^{23}_{10}\text{Ne} \rightarrow ^{23}_{11}\text{Na} + e^- + \bar{\nu} \]
Where:
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