Proton bombardment on `Th^(230)` followed by emission of two alpha particles will produce:

Options

A. `Rn^(232)`
B. `Ra^(233)`
C. `Fr^(223)`
D. `Fr^(222)`

Answer

To solve the problem of what is produced when `Th^(230)` is bombarded with protons and emits two alpha particles, we can follow these steps:

### Step 1: Understand the Initial Reaction
We start with Thorium-230 (`Th^(230)`) and bombard it with a proton (`p`). The atomic number of Thorium is 90, and its mass number is 230.

### Step 2: Write Down the Reaction
The reaction can be represented as:
\[ Th^{230} + p \rightarrow X \]
where \( X \) is the resulting element after the reaction.

### Step 3: Calculate the Mass Number After Proton Bombardment
When a proton is added to Thorium-230, the new mass number becomes:
\[ 230 + 1 = 231 \]
So, we have:
\[ Th^{231} \]

### Step 4: Emission of Two Alpha Particles
Alpha particles are helium nuclei, each with a mass number of 4 and an atomic number of 2. Since two alpha particles are emitted, the total mass number lost is:
\[ 2 \times 4 = 8 \]
The atomic number lost is:
\[ 2 \times 2 = 4 \]

### Step 5: Calculate the Final Mass Number and Atomic Number
Now, we subtract the mass number and atomic number of the emitted alpha particles from the mass number and atomic number of the resulting Thorium-231:
- New mass number:
\[ 231 - 8 = 223 \]
- New atomic number:
\[ 90 - 4 = 86 \]

### Step 6: Identify the Resulting Element
The element with atomic number 86 is Radon (Rn). Therefore, the final product of the reaction is:
\[ Rn^{223} \]

### Final Answer
Thus, the product of proton bombardment on `Th^(230)` followed by the emission of two alpha particles is:
\[ Rn^{223} \]

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