When `.^(30)Si` bombarded with a deutron. `.^(31)Si` is formed in its ground state with the emission of a proton. The energy released in this reaction from the following is `E` then find `[E]`:-

Explanation

To solve the problem of calculating the energy released (E) when silicon-30 is bombarded with a deuteron to form silicon-31 and emit a proton, we can follow these steps:

### Step 1: Write the Nuclear Reaction
The nuclear reaction can be represented as:
\[ \text{ }^{30}\text{Si} + \text{ }^{2}\text{H} \rightarrow \text{ }^{31}\text{Si} + \text{ }^{1}\text{H} \]
Where:
- \(^{30}\text{Si}\) is silicon-30
- \(^{2}\text{H}\) is a deuteron (which consists of one proton and one neutron)
- \(^{31}\text{Si}\) is silicon-31
- \(^{1}\text{H}\) is a proton

### Step 2: Determine the Q-value of the Reaction
The Q-value of the reaction can be calculated using the mass-energy equivalence principle. The Q-value is given by:
\[ Q = (m_{\text{reactants}} - m_{\text{products}})c^2 \]
In terms of masses, we can express it as:
\[ Q = m_{\text{deuteron}} + m_{\text{Si-30}} - m_{\text{Si-31}} - m_{\text{proton}} \]

### Step 3: Use Given Masses and Energy Values
From the problem, we have the following energy values:
- The energy released when silicon-31 decays is \(1.51 \text{ MeV}\).
- The energy released when silicon-30 captures a deuteron is \(5.10 \text{ MeV}\).
- The neutron decay energy is \(0.78 \text{ MeV}\).

### Step 4: Set Up the Equations
We can set up the following equations based on the given information:
1. From the decay of silicon-31:
\[ m_{\text{Si-31}} = m_{\text{proton}} + m_{\text{electron}} + 1.51 \text{ MeV} \]
2. From the reaction involving silicon-30 and deuteron:
\[ m_{\text{Si-30}} + m_{\text{deuteron}} = m_{\text{Si-31}} + m_{\text{neutron}} + 5.10 \text{ MeV} \]

### Step 5: Substitute and Rearrange
Subtract the mass-energy equation of silicon-31 from the reaction equation:
\[ m_{\text{Si-30}} + m_{\text{deuteron}} - m_{\text{Si-31}} = m_{\text{neutron}} + 5.10 - 1.51 \text{ MeV} \]
This simplifies to:
\[ m_{\text{Si-30}} + m_{\text{deuteron}} - m_{\text{Si-31}} = m_{\text{neutron}} + 3.59 \text{ MeV} \]

### Step 6: Relate Neutron Mass to Proton Mass
Using the mass-energy equivalence for the neutron:
\[ m_{\text{neutron}} = m_{\text{proton}} + m_{\text{electron}} + 0.78 \text{ MeV} \]

### Step 7: Final Calculation
Substituting the expression for the neutron mass back into the equation gives:
\[ m_{\text{Si-30}} + m_{\text{deuteron}} - m_{\text{Si-31}} - m_{\text{proton}} = 0 \]
Adding up the energies gives:
\[ 3.59 + 0.78 = 4.37 \text{ MeV} \]

### Conclusion
Thus, the energy released in this reaction is:
\[ E = 4.37 \text{ MeV} \]