Calculate the maximum energy that a `beta` particle can have in the following decay:
`._(8)O^(19) to ._(9)F^(19)+._(-1)e^(0)+barnu`
Given, `m(._(8)O^(19))=19.003576u, m(._(9)F^(19))=18.998403u, m(._(-1)e^(0))=0.000549u`

Explanation

To solve the problem of calculating the maximum energy that a beta particle can have in the decay of \( _{8}O^{19} \) to \( _{9}F^{19} + _{-1}e^{0} + \bar{\nu} \), we will follow these steps:

### Step 1: Identify the masses involved in the decay
We are given the following masses:
- Mass of \( _{8}O^{19} \) (Oxygen-19): \( m( _{8}O^{19} ) = 19.003576 \, u \)
- Mass of \( _{9}F^{19} \) (Fluorine-19): \( m( _{9}F^{19} ) = 18.998403 \, u \)
- Mass of the beta particle (electron): \( m( _{-1}e^{0} ) = 0.000549 \, u \)

### Step 2: Calculate the mass defect (\( \Delta m \))
The mass defect is calculated as the difference between the mass of the reactant and the total mass of the products. The equation for mass defect is:
\[ \Delta m = m( _{8}O^{19} ) - (m( _{9}F^{19} ) + m( _{-1}e^{0} )) \]
Substituting the values:
\[ \Delta m = 19.003576 \, u - (18.998403 \, u + 0.000549 \, u) \]
Calculating this gives:
\[ \Delta m = 19.003576 \, u - 19.000952 \, u = 0.002624 \, u \]

### Step 3: Calculate the Q-value of the reaction
The Q-value, which represents the energy released in the decay, can be calculated using the mass defect and the energy-mass equivalence relation:
\[ Q = \Delta m \cdot c^2 \]
Where \( c^2 \) is given as \( 931.5 \, \text{MeV/u} \). Thus,
\[ Q = 0.002624 \, u \cdot 931.5 \, \text{MeV/u} \]
Calculating this gives:
\[ Q = 2.448 \, \text{MeV} \]

### Step 4: Determine the maximum energy of the beta particle
In beta decay, the maximum energy of the beta particle (electron) is equal to the Q-value of the decay, since the neutrino carries away negligible energy. Therefore, the maximum energy of the beta particle is:
\[ E_{\text{max}} = Q = 2.448 \, \text{MeV} \]

### Final Answer
The maximum energy that the beta particle can have in this decay is approximately **2.448 MeV**.

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