By using the following atomic masses : `._(92)^(238)U = 238.05079u`. `._(2)^(4)He = 4.00260u, ._(90)^(234)Th = 234.04363u`.
`._(1)^(1)H = 1.007834, ._(91)^(237)Pa = 237.065121u` (i) Calculate the energy released during the `alpha-`decay of `._(92)^(238)U`.
(ii) Show that `._(92)^(238)U` cannot spontaneously emit a proton.

Explanation

To solve the problem, we will follow these steps:

### Part (i): Calculate the energy released during the alpha decay of \(_{92}^{238}U\).

1. **Write the decay equation**:
The alpha decay of Uranium-238 can be represented as:
\[ _{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He \]

2. **Identify the atomic masses**:
- Mass of \(_{92}^{238}U = 238.05079 \, u\)
- Mass of \(_{90}^{234}Th = 234.04363 \, u\)
- Mass of \(_{2}^{4}He = 4.00260 \, u\)

3. **Calculate the mass defect (\(\Delta m\))**:
\[ \Delta m = \text{mass of reactants} - \text{mass of products} \]
\[ \Delta m = 238.05079 \, u - (234.04363 \, u + 4.00260 \, u) \]
\[ \Delta m = 238.05079 \, u - 238.04623 \, u = 0.00456 \, u \]

4. **Convert the mass defect to energy**:
Using the conversion factor \(1 \, u = 931.5 \, \text{MeV/c}^2\):
\[ \text{Energy released} = \Delta m \times 931.5 \, \text{MeV/c}^2 \]
\[ \text{Energy released} = 0.00456 \, u \times 931.5 \, \text{MeV/u} \approx 4.24764 \, \text{MeV} \]

### Part (ii): Show that \(_{92}^{238}U\) cannot spontaneously emit a proton.

1. **Write the decay equation for proton emission**:
The reaction for proton emission can be represented as:
\[ _{92}^{238}U \rightarrow _{91}^{237}Pa + _{1}^{1}H \]

2. **Identify the atomic masses**:
- Mass of \(_{91}^{237}Pa = 237.065121 \, u\)
- Mass of \(_{1}^{1}H = 1.007834 \, u\)

3. **Calculate the mass defect (\(\Delta m'\))**:
\[ \Delta m' = \text{mass of reactants} - \text{mass of products} \]
\[ \Delta m' = 238.05079 \, u - (237.065121 \, u + 1.007834 \, u) \]
\[ \Delta m' = 238.05079 \, u - 238.072955 \, u = -0.022165 \, u \]

4. **Interpret the result**:
Since \(\Delta m'\) is negative, it indicates that energy must be supplied to make this reaction occur, meaning that the reaction is not spontaneous.

### Summary of Results:
- **Energy released during alpha decay**: \(4.24764 \, \text{MeV}\)
- **Proton emission**: Not spontaneous due to negative mass defect.