A nuclear transformation is denoted by `X(n,alpha) rarr. _(3)^(7)Li`. Which of the following is the nucleus of element `X`

Options

A. `._(6)^(12)C`
B. `._(5)^(10)B`
C. `._(5)^(9)B`
D. `._(4)^(11)Be`

Explanation

To solve the problem, we need to analyze the nuclear transformation represented by the equation:

\[ X(n, \alpha) \rightarrow \, ^{7}_{3}Li \]

This indicates that nucleus \( X \) is bombarded with a neutron (n) and emits an alpha particle (α), resulting in the formation of lithium-7 (\( ^{7}_{3}Li \)).

### Step-by-step Solution:

1. **Identify the Components of the Reaction:**
- The reaction involves a nucleus \( X \), a neutron (n), and an alpha particle (α).
- The product of the reaction is lithium-7, which has an atomic number (Z) of 3 and a mass number (A) of 7.

2. **Write Down the Conservation Equations:**
- The conservation of nucleons (mass number A) and protons (atomic number Z) must hold true in nuclear reactions.
- Let the mass number and atomic number of nucleus \( X \) be \( A_X \) and \( Z_X \) respectively.

3. **Set Up the Mass Number Equation:**
- The total mass number before the reaction is:
\[ A_X + 1 \] (where 1 is the mass number of the neutron)
- The total mass number after the reaction is:
\[ 7 + 4 \] (where 4 is the mass number of the alpha particle)
- Therefore, we have:
\[ A_X + 1 = 7 + 4 \]
\[ A_X + 1 = 11 \]
\[ A_X = 10 \]

4. **Set Up the Atomic Number Equation:**
- The total atomic number before the reaction is:
\[ Z_X + 0 \] (where 0 is the atomic number of the neutron)
- The total atomic number after the reaction is:
\[ 3 + 2 \] (where 2 is the atomic number of the alpha particle)
- Therefore, we have:
\[ Z_X + 0 = 3 + 2 \]
\[ Z_X = 5 \]

5. **Determine the Element X:**
- Now we have determined that \( A_X = 10 \) and \( Z_X = 5 \).
- The element with atomic number 5 is boron (B), and its mass number is 10, which corresponds to the isotope \( ^{10}_{5}B \).

### Conclusion:
Thus, the nucleus of element \( X \) is \( ^{10}_{5}B \) (Boron-10).

### Final Answer:
The nucleus of element \( X \) is Boron-10 (\( ^{10}_{5}B \)).