In nuclear transformation
`""_(a)X^(b) + ""_(0)n^(1) to ""_(3)Li^(7) + ""_(2)He^(4)`
Which one is the nucleus of X?

Options

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

Answer

To determine the nucleus of X in the nuclear transformation reaction:

\[ \text{ }_{a}X^{b} + \text{ }_{0}n^{1} \rightarrow \text{ }_{3}Li^{7} + \text{ }_{2}He^{4} \]

we will follow these steps:

### Step 1: Identify the components of the reaction
- The reaction involves a nucleus \(X\) with atomic number \(a\) and mass number \(b\), a neutron (which has atomic number 0 and mass number 1), lithium with atomic number 3 and mass number 7, and an alpha particle (helium) with atomic number 2 and mass number 4.

### Step 2: Balance the atomic numbers
- On the left side (reactants), the total atomic number is \(a + 0 = a\).
- On the right side (products), the total atomic number is \(3 + 2 = 5\).
- Therefore, we can set up the equation:
\[ a = 5 \]

### Step 3: Balance the mass numbers
- On the left side, the total mass number is \(b + 1\) (since the neutron contributes 1).
- On the right side, the total mass number is \(7 + 4 = 11\).
- Thus, we can set up the equation:
\[ b + 1 = 11 \]
- Solving for \(b\):
\[ b = 11 - 1 = 10 \]

### Step 4: Identify the nucleus of X
- Now we have determined that \(a = 5\) and \(b = 10\).
- The nucleus \(X\) is characterized by atomic number 5 and mass number 10.
- Referring to the periodic table, the element with atomic number 5 is boron (B).

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

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