Consider the following nuclear decay: `""_(92)^(236)Uto""_(90)^(232)Th+X`
What is X?

Options

A. `alpha` decay
B. `p`
C. `beta^(+)`
D. `beta^(-)`

Answer

To solve the nuclear decay equation given by:

\[ _{92}^{236}\text{U} \rightarrow _{90}^{232}\text{Th} + X \]

we need to identify what \(X\) is.

### Step 1: Analyze the decay process
In nuclear decay, the total number of nucleons (protons + neutrons) and the total number of protons must be conserved.

### Step 2: Write down the known values
- Uranium-236 (\( _{92}^{236}\text{U} \)) has:
- Atomic number (protons) = 92
- Mass number (nucleons) = 236

- Thorium-232 (\( _{90}^{232}\text{Th} \)) has:
- Atomic number (protons) = 90
- Mass number (nucleons) = 232

### Step 3: Calculate the missing values for \(X\)
1. **Conservation of Protons:**
\[ \text{Protons in } U = \text{Protons in } Th + \text{Protons in } X \]
\[ 92 = 90 + \text{Protons in } X \]
Therefore,
\[ \text{Protons in } X = 92 - 90 = 2 \]

2. **Conservation of Nucleons:**
\[ \text{Nucleons in } U = \text{Nucleons in } Th + \text{Nucleons in } X \]
\[ 236 = 232 + \text{Nucleons in } X \]
Therefore,
\[ \text{Nucleons in } X = 236 - 232 = 4 \]

### Step 4: Identify \(X\)
Since \(X\) has 2 protons and a total of 4 nucleons, it must have 2 neutrons as well (since \(4 - 2 = 2\)). This corresponds to the Helium nucleus, which is commonly referred to as an alpha particle.

### Conclusion
Thus, \(X\) is an alpha particle, which can be denoted as:

\[ X = _{2}^{4}\text{He} \text{ or } \alpha \]

### Final Answer
The value of \(X\) is an alpha particle.

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