In compound `X(n, alpha) rarr ._(3)Li^(7)`, the element `X` is

Options

A. `""_(2)He^(4)`
B. `""_(5)B^(10)`
C. `""_(5)B^(9)`
D. `""_(4)Be^(11)`

Answer

To solve the problem, we need to identify the element \( X \) in the nuclear reaction:

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

### Step 1: Understand the Reaction
In this reaction, a nucleus \( X \) captures a neutron (n) and emits an alpha particle (α), resulting in the formation of lithium-7 (\( _{3}^{7}\text{Li} \)).

### Step 2: Write the General Form of the Reaction
The general form of the reaction can be expressed as:

\[ _{Z}^{A}X + n \rightarrow _{3}^{7}\text{Li} + _{2}^{4}\alpha \]

Where:
- \( Z \) is the atomic number of element \( X \)
- \( A \) is the mass number of element \( X \)

### Step 3: Conservation of Mass Number
We apply the conservation of mass number (A):

\[ A + 1 = 7 + 4 \]

Here, \( A + 1 \) accounts for the neutron added to nucleus \( X \), and \( 7 + 4 \) accounts for the lithium nucleus and the emitted alpha particle.

Calculating gives:

\[ A + 1 = 11 \implies A = 10 \]

### Step 4: Conservation of Atomic Number
Next, we apply the conservation of atomic number (Z):

\[ Z + 0 = 3 + 2 \]

Here, \( Z \) is the atomic number of \( X \), \( 3 \) is the atomic number of lithium, and \( 2 \) is the atomic number of the alpha particle.

Calculating gives:

\[ Z = 5 \]

### Step 5: Identify the Element
Now we have \( Z = 5 \) and \( A = 10 \). The element with atomic number 5 is boron (\( B \)).

Thus, the element \( X \) is:

\[ _{5}^{10}\text{B} \]

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

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