Find energy released in the alpha decay,
Given `_92^238Urarr_90^234Th+_2^4He`
`M(`_92^238U)=238.050784u`
`M(`_90^234Th)=234.043593u`
`M(`_2^4He)=4.002602u`

Explanation

To find the energy released in the alpha decay of Uranium-238, we can follow these steps:

### Step 1: Identify the reaction
The decay reaction is given as:
\[ _{92}^{238}\text{U} \rightarrow _{90}^{234}\text{Th} + _{2}^{4}\text{He} \]

### Step 2: Write down the masses
The masses of the isotopes involved in the reaction are:
- Mass of Uranium-238, \( M(^{238}\text{U}) = 238.050784 \, u \)
- Mass of Thorium-234, \( M(^{234}\text{Th}) = 234.043593 \, u \)
- Mass of Helium-4, \( M(^{4}\text{He}) = 4.002602 \, u \)

### Step 3: Calculate the total mass of reactants
The total mass of the reactants (Uranium-238) is:
\[ M_{\text{reactants}} = M(^{238}\text{U}) = 238.050784 \, u \]

### Step 4: Calculate the total mass of products
The total mass of the products (Thorium-234 and Helium-4) is:
\[ M_{\text{products}} = M(^{234}\text{Th}) + M(^{4}\text{He}) = 234.043593 \, u + 4.002602 \, u = 238.046195 \, u \]

### Step 5: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the mass of the reactants and the mass of the products:
\[ \Delta m = M_{\text{reactants}} - M_{\text{products}} = 238.050784 \, u - 238.046195 \, u = 0.004589 \, u \]

### Step 6: Convert mass defect to energy
Using Einstein's equation \( E = \Delta mc^2 \), we need to convert the mass defect from atomic mass units (u) to energy in MeV. The conversion factor is:
\[ 1 \, u \approx 931.5 \, \text{MeV/c}^2 \]
Thus, the energy released (\( E \)) is:
\[ E = \Delta m \cdot 931.5 \, \text{MeV/u} = 0.004589 \, u \cdot 931.5 \, \text{MeV/u} \approx 4.27 \, \text{MeV} \]

### Final Answer
The energy released in the alpha decay of Uranium-238 is approximately:
\[ \boxed{4.27 \, \text{MeV}} \]

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