If `._(92)U^(235)` assumed to decay only by emitting two `alpha`-and one `beta`-particles, the possible product of decays is

Options

A. `._(89)Ac^(231)`
B. `._(89)Ac^(235)`
C. `._(89)Ac^(236)`
D. `._(89)Ac^(227)`

Explanation

To solve the problem of determining the decay products of Uranium-235 (U-235) when it emits two alpha particles and one beta particle, we can follow these steps:

### Step 1: Understand the decay process
- **Alpha decay**: When a nucleus emits an alpha particle (which is essentially a helium nucleus, \( _{2}^{4}\text{He} \)), it loses 2 units of atomic number and 4 units of mass number.
- **Beta decay**: When a nucleus emits a beta particle (which is an electron), it loses 1 unit of atomic number but does not change its mass number.

### Step 2: Write down the initial values
- For Uranium-235, we have:
- Atomic number (Z) = 92
- Mass number (A) = 235

### Step 3: Calculate the changes due to the emissions
- **For two alpha particles**:
- Change in atomic number = \( 2 \times 2 = 4 \) (decrease)
- Change in mass number = \( 2 \times 4 = 8 \) (decrease)

- **For one beta particle**:
- Change in atomic number = \( 1 \) (decrease)
- Change in mass number = \( 0 \) (no change)

### Step 4: Total changes in atomic and mass numbers
- Total change in atomic number:
\[ \text{Total decrease in Z} = 4 + 1 = 5 \]
- Total change in mass number:
\[ \text{Total decrease in A} = 8 + 0 = 8 \]

### Step 5: Calculate the final atomic and mass numbers
- Final atomic number:
\[ Z = 92 - 5 = 87 \]
- Final mass number:
\[ A = 235 - 8 = 227 \]

### Step 6: Identify the new element
- The element with atomic number 87 is **Francium (Fr)**. Therefore, the final product after the decay of Uranium-235, emitting two alpha particles and one beta particle, is:
\[ _{87}^{227}\text{Fr} \]

### Final Answer
The possible product of the decays is **Francium-227**.

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