What will be the product of reaction `._(101)^(255)Md(alpha,2n)`?

Options

A. `._(103)^(256)Lr`
B. `._(54)^(257)No`
C. `._(103)^(257)Lr`
D. `._(82)^(205)Pb`

Answer

To solve the reaction \( _{101}^{255}Md + \alpha \rightarrow X + 2n \), we need to determine the product \( X \) formed when mendelevium (Md) reacts with an alpha particle and releases two neutrons.

### Step-by-Step Solution:

1. **Identify the Reactants:**
- Mendelevium (Md) has an atomic number of 101 and a mass number of 255.
- An alpha particle (\( \alpha \)) is a helium nucleus with an atomic number of 2 and a mass number of 4.

2. **Write the Reaction:**
The reaction can be represented as:
\[ _{101}^{255}Md + _{2}^{4}\alpha \rightarrow X + 2n \]
where \( n \) represents neutrons.

3. **Calculate Total Atomic and Mass Numbers on the Left Side:**
- Total atomic number on the left:
\[ 101 + 2 = 103 \]
- Total mass number on the left:
\[ 255 + 4 = 259 \]

4. **Determine the Atomic and Mass Numbers of Product \( X \):**
Since two neutrons are released, they do not contribute to the atomic number but do contribute to the mass number.
- The atomic number of \( X \) will be:
\[ 103 \quad (\text{since neutrons have no charge}) \]
- The mass number of \( X \) will be:
\[ 259 - 2 = 257 \]

5. **Identify the Product \( X \):**
The product \( X \) has an atomic number of 103 and a mass number of 257. The element with atomic number 103 is Lawrencium (Lr).

6. **Final Reaction Representation:**
The balanced reaction can be written as:
\[ _{101}^{255}Md + _{2}^{4}\alpha \rightarrow _{103}^{257}Lr + 2n \]

### Conclusion:
The product of the reaction \( _{101}^{255}Md + \alpha \rightarrow X + 2n \) is \( _{103}^{257}Lr \) (Lawrencium).