When helium nuclei bombard beryllium nuclei ,then

Options

A. Electrons are emitted
B. Protons are emitted
C. Neutrons are emitted
D. protons and neutrons are emitted

Answer

To solve the problem of what happens when helium nuclei bombard beryllium nuclei, we can follow these steps:

### Step 1: Identify the Nuclei Involved
- Helium nucleus (He) has an atomic number (Z) of 2 and a mass number (A) of 4, represented as \( ^4_2He \).
- Beryllium nucleus (Be) has an atomic number (Z) of 4 and a mass number (A) of 9, represented as \( ^9_4Be \).

### Step 2: Write the Reaction Equation
When helium nuclei bombard beryllium nuclei, we can represent the reaction as:
\[ ^4_2He + ^9_4Be \rightarrow ? \]

### Step 3: Determine the Resulting Nucleus
- The total number of protons (atomic number) before the reaction is \( 2 + 4 = 6 \).
- The total mass number before the reaction is \( 4 + 9 = 13 \).
- The resulting nucleus will have an atomic number of 6, which corresponds to carbon (C). The most stable isotope of carbon is \( ^{12}_6C \).

### Step 4: Balance the Mass Number
- On the right side, we have \( ^{12}_6C \) with a mass number of 12.
- Since the initial mass number was 13, we need to account for the missing nucleon:
\[ 13 - 12 = 1 \]
This indicates that one nucleon is emitted.

### Step 5: Determine the Type of Emitted Particle
- The atomic number (number of protons) is conserved, meaning no protons are emitted.
- Since the mass number decreases by 1 and the number of protons remains the same, the emitted particle must be a neutron.

### Conclusion
Thus, when helium nuclei bombard beryllium nuclei, one neutron is emitted in the process.

### Final Reaction Equation
The complete equation can be summarized as:
\[ ^4_2He + ^9_4Be \rightarrow ^{12}_6C + ^1_0n \]
where \( ^1_0n \) represents the emitted neutron.

### Answer
The particle emitted during this process is a neutron.

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