When `_(3)Li^(7)` nuclei are bombarded by protons , and the resultant nuclei are `_(4)Be^(8)` , the emitted particle will be

Options

A. alpha particles
B. beta particles
C. gamma photons
D. neutrons

Answer

To solve the problem, we need to analyze the nuclear reaction involving lithium nuclei when they are bombarded by protons. Let's break it down step by step.

### Step 1: Identify the Reactants
The reactants in this nuclear reaction are:
- Lithium-7 nuclei, denoted as \(_{3}^{7}\text{Li}\)
- Protons, denoted as \(_{1}^{1}\text{H}\)

### Step 2: Write the Reaction
When a lithium-7 nucleus is bombarded by a proton, the reaction can be represented as:
\[ _{3}^{7}\text{Li} + _{1}^{1}\text{H} \rightarrow _{4}^{8}\text{Be} + X \]
where \(X\) is the emitted particle that we need to determine.

### Step 3: Determine the Products
From the reaction, we see that the resultant nucleus is Beryllium-8, denoted as \(_{4}^{8}\text{Be}\).

### Step 4: Check Conservation of Mass and Atomic Numbers
Now, we need to ensure that both mass number and atomic number are conserved in this reaction.

- **Mass Number**:
- Before the reaction: \(7 + 1 = 8\)
- After the reaction: Mass number of Beryllium-8 is \(8\) and the mass number of the emitted particle \(X\) will be \(m_X\).

Therefore, we have:
\[ 8 = 8 + m_X \implies m_X = 0 \]

- **Atomic Number**:
- Before the reaction: \(3 + 1 = 4\)
- After the reaction: Atomic number of Beryllium is \(4\) and the atomic number of the emitted particle \(X\) will be \(z_X\).

Therefore, we have:
\[ 4 = 4 + z_X \implies z_X = 0 \]

### Step 5: Identify the Emitted Particle
The emitted particle \(X\) has both a mass number of \(0\) and an atomic number of \(0\). The only particle that fits this description is a gamma particle, denoted as \(\gamma\).

### Conclusion
Thus, the emitted particle when \(_{3}^{7}\text{Li}\) nuclei are bombarded by protons resulting in \(_{4}^{8}\text{Be}\) is a gamma particle.

### Final Answer
The emitted particle will be \(\gamma\) (gamma particle).

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